MATH 409 Lecture 23

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Lecture Slides

Infinite Series

Given a sequence $\left\{a_{n}\right\}$ of real numbers, the expression

$a_{1}+a_{2}+\dots =\sum _{n=1}^{\infty }$

Is called an infinite series with terms $a_{n}$

The partial sum of order $n$ is given by

$s_{n}=a_{1}+\dots +a_{n}$

If the sequence $\left\{s_{n}\right\}$ converges to limit $s\in \mathbb {R}$, we say the series converges to $s$ or that $s$ is the sum of the series and write $\sum _{n=1}^{\infty }a_{n}=s$

Otherwise the series diverges

Cauchy Criterion

Theorem. [Cauchy Criterion]. An infinite series $\sum _{n=1}^{\infty }a_{n}$ converges if and only if for every $\epsilon >0$ there exists $N\in \mathbb {N}$ such that $m\geq n\geq N$ implies $\left|a_{n}+a_{n+1}+\dots +a_{m}\right|<\epsilon$

Proof. Let $\left\{s_{n}\right\}$ be the sequence of partial sums. Then

$a_{n}+a_{n}+1+\dots +a_{m}=s_{m}-s_{n-1}$

Consequently, the condition of the theorem is equivalent to the condition that $\left\{s_{n}\right\}$ be a Cauchy sequence. As we know, a sequence is convergent if and only if it is a Cauchy sequence.

quod erat demonstrandum

Examples

${\frac {1}{2}}+{\frac {1}{2^{2}}}+\dots +{\frac {1}{2^{n}}}+\dots =1$

The partial sums $s_{n}$ of this series satisfy $s_{n}=1-2^{-n}$ for all $n\in \mathbb {N}$. Thus $s_{n}\to 1$ as $n\to \infty$

${\frac {1}{1\cdot 2}}+{\frac {1}{2\cdot 3}}+\dots +{\frac {1}{n(n+1)}}+\dots =1$

Since ${\frac {1}{n(n+1)}}={\frac {1}{n}}-{\frac {1}{n+1}}$, the partial sums $s_{n}$ of this series satisfy $s_{n}=1-{\frac {1}{n+1}}$. Thus $s_{n}\to 1$ as $n\to \infty$

$\sum _{n=1}^{\infty }\left(-1\right)^{n}=-1+1-1+\dots$ diverges

The partial sums $s_{n}$ satisfy $s_{n}=-1$ for odd $n$ and $s_{n}=0$ for even $n$. Hence the sequence $\left\{s_{n}\right\}$ has no limit.

The geometric series $\sum _{n=0}^{\infty }x^{n}$ converges if and only if $\left|x\right|<1$, in which case its sum is ${\frac {1}{1-x}}$.

In the case $\left|x\right|\geq 1$, the series fails the divergence test. For any $x\neq 1$, the partial sums of the geometric series satisfy

$s_{n}=1+x+x^{2}+\dots +x^{n}={\frac {1-x^{n+1}}{1-x}}$

For $\left|x\right|<1$, $s_{n}\to {\frac {1}{1-x}}$ as $n\to \infty$.

Properties of Infinite Series

Theorem. [Divergence Test]. If the terms of an infinite series do not converge to zero, then the series diverges.

Theorem. [Linearity]. If $\sum _{n=1}^{\infty }a_{n}$ and $\sum _{n=1}^{\infty }b_{n}$ are convergent series, then

$\sum _{n=1}^{\infty }(a_{n}+b_{n})=\sum _{n=1}^{\infty }a_{n}+\sum _{n=1}^{\infty }b_{n}$

and

$\sum _{n=1}^{\infty }\left(r\,a_{n}\right)=r\,\sum _{n=1}^{\infty }a_{n}$

for any $r\in \mathbb {R}$.

Theorem. If $\sum _{n=1}^{\infty }a_{n}$ and $\sum _{n=1}^{\infty }b_{n}$ are convergent series, and $a_{n}\leq b_{n}$ for all $n\in \mathbb {N}$, then

$\sum _{n=1}^{\infty }a_{n}\leq \sum _{n=1}^{\infty }b_{n}$

Series with Nonnegative Terms

Suppose that a series $\sum _{n=1}^{\infty }a_{n}$ has nonnegative terms $a_{n}\geq 0$ for all $n\in \mathbb {N}$. Then the sequence of partial sums $s_{n}=a_{1}+\dots +a_{n}$ is increasing. It follows that $\left\{s_{n}\right\}$

• converges to a finite limit if bounded and
• diverges to $+\infty$ otherwise.

In the latter case, we write $\sum _{n=1}^{\infty }a_{n}=\infty$.

Comparison Test

Theorem. [Comparison Test]. Suppose that $a_{n},b_{n}\geq 0$ for all $n\in \mathbb {N}$ and $a_{n}\leq b_{n}$ for large enough $n$. Then

• convergence of the series $\sum _{n=1}^{\infty }b_{n}$ (larger terms) implies convergence of $\sum _{n=1}^{\infty }a_{n}$ (smaller terms), while
• divergence of the series $\sum _{n=1}^{\infty }a_{n}=\infty$ (smaller terms) implies divergence of $\sum _{n=1}^{\infty }b_{n}=\infty$ (larger terms)

Proof. Since change a finite number of terms does not affect convergence of a series, it is no loss to assume that $a_{n}\leq b_{n}$ for all $n\in \mathbb {N}$. Then the partial sums $s_{n}=\sum _{k=1}^{n}a_{k}$ and $t_{n}=\sum _{k=1}^{n}b_{n}$ satisfy $s_{n}\leq t_{n}$ for all $n$. Consequently, if $s_{n}\to +\infty$ as $n\to \infty$, then also $t_{n}\to +\infty$ as $n\to \infty$.

Conversely, if $\left\{t_{n}\right\}$ is bounded, then so is $\left\{s_{n}\right\}$.

Integral Test

Theorem. [Integral test]. Suppose a function $f:\left[1,\infty \right)\to \mathbb {R}$ is positive and decreasing on $\left[1,\infty \right)$. Then

1. a sequence $\left\{y_{n}\right\}$ is bounded, where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_n = f(1) + f(2) + \dots + f(n) - \int_1^n f(x) \,\mathrm{d}x} $n=1,2,\ldots$
2. the series $\sum _{n=1}^{\infty }f(n)$ is convergent if and only if the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is improperly integrable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ 1, \infty \right)} .

Proof. To prove the theorem, we need the following lemma:

Lemma. Any monotone function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g : \left[ a,b \right] \to \mathbb{R}} is integrable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ a,b \right]} .

Idea of the proof. Any monotone function has only jump discontinuities. Further, any function has at most countably many jump discontinuities. Besides, a monotone function on a closed interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ a,b \right]} is clearly bounded.

The lemma implies that the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is integrable on every closed interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J = [a,b] \subset \left[ 1, \infty \right)} . Then for any partition Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P} of the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J} , the lower Darboux sum Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L(f,P)} and the upper Darboux sum Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U(f,P)} satisfy

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L(f,P) \le \int_a^b f(x) \,\mathrm{d}x \le U(f,P)}

Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P = \left\{ x_0, x_1, \ldots, x_k \right\}} , where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0 < x_1 < \dots < x_k} . Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sup{f\left( \left[ x_{j-1}, x_j \right] \right)} = f(x_{j-1})} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \inf{f\left( \left[ x_{j-1}, x_j \right] \right)} = f(x_j)} since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is decreasing. In the case Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J = \left[ 1,n \right]} , where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \in \mathbb{N}} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P = \left\{ 1, 2, \ldots, n \right\}} , we obtain

• Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L(f,P) = f(2) + f(3) + \dots + f(n)} ,
• Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U(f,P) = f(1) + f(2) + \dots + f(n-1)} .

Then the above inequalities imply that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 < f(n) \le y_n \le f(1)} . Thus the sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ y_n \right\}} is bounded.

Now for the second part of the theorem: Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is positive, the series Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty f(n)} either converges or else it diverges to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle +\infty} . Likewise, the improper integral Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_1^\infty f(x) \,\mathrm{d}x} either converges or else it diverges to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle +\infty} . Since the sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\{ y_n \right\}} is bounded by the above, divergence of the series and the integral imply each other.

quod erat demonstrandum

Examples

Theorem. [P-Series Test]. [Riemann Zeta Function]. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty \frac{1}{n^p}} is convergent for any Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p > 1} and divergent for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p < 1} .

Proof. For any Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p \ne 1} , we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^{-p} \,\mathrm{d}x = \frac{x^{1-p}}{1-p} + C} on the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ 1, \infty \right)} . The antiderivative converges to a finite limit at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle +\infty} in the case Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p > 1} and diverges to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle +\infty} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p < 1} . Hence the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = x^{-p}} is improperly integrable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ 1, \infty \right)} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p > 1} , but not for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p < 1} . By the Integral Test, the series is convergent for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p > 1} and divergent for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 \le p < 1} .

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p < 0} then the Integral Test does not apply since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is not decreasing. In this case, the series is divergent since the terms Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{n^p}} do not converge to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0} as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \to \infty} .

quod erat demonstrandum

Harmonic Series. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty \frac{1}{n}} diverges.

Indeed Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_1^n \frac{1}{x} \,\mathrm{d}x = \log{x} \to +\infty} as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \to 0} . By the integral test, the series is divergent.

Moreover, the sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_n = \sum_{k=1}^n k^{-1} - \log{n}} is bounded (actually, it is decreasing and hence convergent)

This was the bonus problem on the test.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=2}^\infty \frac{1}{n \log^2{n}}} converges.

The antiderivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = \left( x \, \log^2{x} \right)^{-1}} on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1, \infty)} is

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{1}{x \, \log^2{x}} \,\mathrm{d}x = -\frac{1}{\log{x}} + C}

Since the antiderivative converges to a finite limit at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle +\infty} , the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is improperly integrable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ 2, \infty \right)} .

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty \frac{1}{1+n^2}} converges.

Indeed, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 < \frac{1}{1+n^2} \le \frac{1}{n^2}} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \in \mathbb{N}} . Since the series Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}} is convergent, it remains to apply the comparison test. Alternatively, we can use the integral test.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{1}{1+x^2} \,\mathrm{d}x = \arctan{x} + C} converges to a finite limit (Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\pi}{2} + C} ) at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle +\infty} so the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = \frac{1}{1+x^2}} is improperly integrable on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ 1,\infty \right)} .

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty \mathrm{e}^{-n^2}} converges (really fast!)

We have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 < \mathrm{e}^{-n^2} \le \mathrm{e}^{-n}} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \in \mathbb{N}} . The geometric series Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty \mathrm{e}^{-n} = \sum_{n=1}^\infty \left( \frac{1}{e} \right)^n} is convergent since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 < \mathrm{e}^{-1} < 1} . By the comparison test, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty \mathrm{e}^{-n^2}} is convergent as well.