MATH 152 Chapter 10.3

End Exam 2 Content

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For all of these sum convergence tests to work, all term values must be positive.

Integral Test

Conditions: $a_{n}=f(n)$: $f$ must be continuous, positive, and decreasing

If $\textstyle {\int _{1}^{\infty }{f(n)\,dn}}$ diverges/converges, then series is also divergent/convergent.

Proof: Riemann Sum

${\begin{aligned}{\mbox{left endpoint}}>&\int _{1}^{\infty }{f(x)\,dx}\\{\mbox{right endpoint}}<&\int _{1}^{\infty }{f(x)\,dx}\end{aligned}}$

P-Test

Find the values of $p$ for which $\textstyle \sum _{n=1}^{\infty }{\frac {1}{n^{p}}}$ is convergent.

If p ≤ 0, $\lim _{n\to \infty }a_{n}\nrightarrow 0$

If p > 0, $f(x)={\tfrac {1}{x^{p}}}$ is continuous, positive, and decreasing $\left(f'(x)<0\quad p\neq 1\right)$

$\sum _{n=1}^{\infty }{\frac {1}{n^{p}}}\;{\mbox{converges iff}}\;p>1$

Remainder (Error) Formula

$\int _{N+1}^{\infty }{f(x)\,dx}\leq s-s_{N}\leq \int _{N}^{\infty }{f(x)\,dx}$

Comparison Test

If $a_{n}$ and $b_{n}$ are convergent and $0\leq {a_{n}}\leq {b_{n}}$ for all $n(>N)$:

1. if $\textstyle \sum {b_{n}}$ is convergent, then $\textstyle \sum {a_{n}}$ is also convergent
2. if $\textstyle \sum {a_{n}}$ is divergent, then $\textstyle \sum {b_{n}}$ is also divergent

Example

$\sum _{n=1}^{\infty }{\frac {\sin {\left({\frac {1}{n}}\right)}}{1+n^{2}}}$

Compare to ${\tfrac {1}{n^{2}}}$ because we know that $\sin {x}\leq 1$. We also know that $n^{2}+1>n^{2}$, so ${\tfrac {1}{n^{2}+1}}<{\tfrac {1}{n^{2}}}$

We know that $\textstyle \sum _{n=1}^{\infty }{\tfrac {1}{n^{2}}}$ converges by p-test, and ${\tfrac {\sin {\left({\tfrac {1}{n}}\right)}}{1+n^{2}}}\leq {\tfrac {1}{n^{2}}}$; so...

$\sum _{n=0}^{\infty }{\frac {\sin {\left({\frac {1}{n}}\right)}}{1+n^{2}}}\;{\mbox{must be convergent by comparison with}}\;\sum _{n=0}^{\infty }{\frac {1}{n^{2}}}$

Friday, October 22, 2010 — We know nothing!

Limit Comparison Test

If $a_{n}>0$ and $b_{n}>0$ and $\textstyle {\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}>0}$ (in other words, if $a_{n}$ and $b_{n}$ behave similarly), then $\textstyle \sum {a_{n}}$ and $\textstyle \sum {b_{n}}$ must either both converge or both diverge.

Example 1 (straight comparison)

$\sum _{n=0}^{\infty }{\frac {n+n^{2}}{1+n+n^{3}}}$

$\textstyle \lim _{n\to \infty }\to 0$ If terms approach zero, we know NOTHING! Integration is a last resort, so let's try comparing it with $\textstyle \sum {\frac {1}{n}}$:

We know that $\textstyle \sum {\frac {1}{n}}$ diverges, so we want the other sum to be "pushed up" by this known sum:

${\begin{aligned}{\frac {n+n^{2}}{1+n+n^{3}}}&>^{?}{\frac {1}{n}}\\n^{2}+n^{3}&>^{?}1+n+n^{3}\\n^{2}&>1+n\end{aligned}}$

Therefore, $\textstyle \sum {\frac {n+n^{2}}{1+n+n^{3}}}$ diverges by comparison with $\textstyle \sum {\frac {1}{n}}$.

Example 2 (limit comparison)

What about $\textstyle \sum _{n=1}^{\infty }{\frac {n+n^{2}}{1+n^{2}+n^{3}}}$? Let's try comparing it with $\textstyle \sum {\frac {1}{n}}$ again:

${\begin{aligned}{\frac {n+n^{2}}{1+n^{2}+n^{3}}}&>^{?}{\frac {1}{n}}\\n^{2}+n^{3}&>^{?}1+n^{2}+n^{3}\\0&\ngtr 1\end{aligned}}$

Let's try the Limit Comparison Test (LCT): Let $a_{n}={\tfrac {n+n^{2}}{1+n^{2}+n^{3}}}\;\left(>0\right)$ and $b_{n}={\tfrac {1}{n}}\;\left(>0\right)$

${\begin{aligned}\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=&\lim _{n\to \infty }{\frac {\frac {n+n^{2}}{1+n^{2}+n^{3}}}{\frac {1}{n}}}\\=&\lim _{n\to \infty }{\frac {n^{2}+n^{3}}{1+n^{2}+n^{3}}}\\=&1\end{aligned}}$

Therefore, $a_{n}$ and $b_{n}$ must both be convergent or divergent. Since we know that $\textstyle \sum {\frac {1}{n}}$ is divergent, $\textstyle \sum {\frac {n+n^{2}}{1+n^{2}+n^{3}}}$ must also be divergent.

$\sum {\frac {n+n^{2}}{1+n^{2}+n^{3}}}\;{\mbox{is divergent by LCT with}}\;\sum {\frac {1}{n}}$