MATH 152 Chapter 10.4

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Other Tests for Convergence

What if not all terms are positive? $\left({\mbox{i.e.}}\;\textstyle \sum \left(-1\right)^{n}a_{n}\right)$

Alternating Series Test

Given $a_{n}$, which is positive, decreasing, and approaching zero, then the series $\textstyle \sum \left(-1\right)^{n}a_{n}$ is convergent.

Alternating Harmonic Series (as compared to regular harmonic series) is convergent:

$\sum _{n=1}^{\infty }{\frac {\left(-1\right)^{n+1}}{n}},\quad a_{n}={\frac {1}{n}}$

$a_{n}$ is positive, decreasing, and approaching zero, so the alternating series is convergent by the Alternating Series Test.

As $n$ increases, we pass $s$, the sum of the series in either direction, but get closer to it each time.

Error Test for AST

If $\textstyle \sum \left(-1\right)^{n}a_{n}$ is convergent by AST, then:

$\left|s-s_{N}\right|<a_{N+1}$

Example: Approximating Alternating Sums

Find $\sum _{n=1}^{\infty }{\frac {\left(-1\right)^{n+1}}{n^{8}}}$ to within 4 decimal places.

$a_{n}={\frac {1}{n^{8}}}$, error is less than .0001

${\begin{aligned}\left|s-s_{N}\right|&<a_{N+1}<.0001\\{\frac {1}{\left(N+1\right)^{8}}}&<{\frac {1}{10000}}\\\left(N+1\right)^{8}&>10000\\N+1&>{\sqrt[{8}]{10000}}\\N&>{\sqrt[{8}]{10000}}-1\\N&>2.1623\end{aligned}}$

Let $N=3$ for our approximation:

${\begin{aligned}\sum _{n=1}^{3}{\frac {\left(-1\right)^{n+1}}{n^{8}}}&={\frac {1}{1^{8}}}-{\frac {1}{2^{8}}}+{\frac {1}{3^{8}}}\\&=1-{\frac {1}{256}}+{\frac {1}{6561}}\\&={\frac {1\,673\,311}{1\,679\,616}}\approx 0.9962\end{aligned}}$

Absolutely Convergent

$\textstyle \sum \left|a_{n}\right|$ is convergent, then $\textstyle \sum a_{n}$ is absolutely convergent. Also, if $\textstyle \sum a_{n}$ is absolutely convergent, then $\textstyle \sum a_{n}$ is convergent.

Example: is $\sum _{n=1}^{\infty }{\frac {\left(-1\right)^{n+1}}{n}}$ absolutely convergent?

No, $\sum \left|a_{n}\right|=\sum {\frac {1}{n}}$, which is divergent

However, $\sum _{n=1}^{\infty }{\frac {\left(-1\right)^{n+1}}{n^{2}}}$ is absolutely convergent. (AST + P-Test)

Conditionally Convergent

If a sum converges by Alternating Series Test, but is not Absolutely Convergent, it is conditionally convergent

$\textstyle \sum _{n=1}^{\infty }{\frac {\left(-1\right)^{n+1}}{n}}$ is conditionally convergent.

Ratio Test

Similar to a geometric series (in that $n$ is in the exponent, and $r$ is between 0 and 1). Three Outcomes:

1. If $\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|<1$, then $\textstyle \sum {a_{n}}$ is absolutely convergent
2. If $\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|>1$, then $\textstyle \sum {a_{n}}$ is divergent
3. If $\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|=1$, then WE KNOW NOTHING!

Example

$\sum _{n=1}^{\infty }{\frac {1000^{n}}{n!}}$

${\begin{aligned}\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|=&{\frac {\frac {\left(1000\right)^{n+1}}{\left(n+1\right)!}}{\frac {\left(1000\right)^{n}}{n}}}\\=&\lim _{n\to \infty }{\frac {\left(1000\right)^{n+1}}{\left(n+1\right)n!}}\,{\frac {n!}{\left(1000\right)^{n}}}\\=&\lim _{n\to \infty }{\frac {1000}{n+1}}\\=&0<1\end{aligned}}$

Therefore, $\sum _{n=1}^{\infty }{\frac {1000^{n}}{n!}}$ must be absolutely convergent by the ratio test, and therefore $a_{n}\to {\frac {1000^{n}}{n!}}$ by test for divergence.

Guidelines for Testing

Chapter 10 Review (mixed problems; practice applying tests

(See Series Tests→)

If $a_{n}$ ...	then try ...
does not approach zero	Test for Divergence
has factorials or exponentials $\left(e^{n}\right)$	Ratio Test
alternates signs; has $\left(-1\right)^{n}$	Alternating Series Test, then test $\textstyle \sum \left|a_{n}\right|$ for absolute convergence.
involves fractions with powers of $n$, like $n^{2}$, $n^{3}$, $n^{4}$, etc.	Limit Comparison Test or Straight Comparison.
comes from a decreasing, easy-to-integrate function	Integral Test