## Alternating Series Test

We have discussed whether a series $\sum a_{n}$ will converge, when $a_{n}\geq0$ for each $n$. But what about when we let $a_{n}$ be anything?

What happens to $\sum (-1)^{n}/n$? When we expand it out, we see that it looks like $(1/n)-(1/(n+1)) = 1/(n^{2}+n)$. But by comparison $1/(n^{2}+n)\leq 1/n^{2}$ for each $n>0$. And we know the series $\sum 1/n^{2}$ converges while $\sum 1/n$ diverges. So what happens to $\sum (-1)^{n}/n$?

1. Proposition. Let
(1)$\displaystyle \sum^{\infty}_{n=1}(-1)^{n+1}a_{n} = a_{1}-a_{2}+a_{3}-a_{4}+\dots$
be a given series where ${a_{n}>0}$. Then the series converges if

1. ${a_{n}\geq a_{n+1}}$ for each ${n}$;
2. ${\displaystyle\lim_{n\rightarrow\infty}a_{n}=0}$.

Proof: We have in the sequence of partial sums
(2)$\displaystyle S_{2n}=\underbrace{(a_{1}-a_{2})}_{\geq0} +\underbrace{(a_{3}-a_{4})}_{\geq0}+\dots+\underbrace{(a_{2n-1}-a_{2n})}_{\geq0}\geq0$
so
(3)$\displaystyle 0\leq S_{2}\leq S_{4}\leq \dots\leq S_{2n}\leq\dots$
But we also have
(4)\displaystyle \begin{aligned} S_{2n} &= a_{1} - (a_{2}-a_{3})-(a_{4}-a_{5})-(\dots)-a_{2n}\\ &\leq a_{1}. \end{aligned}
So
(5)$\displaystyle \lim_{n\rightarrow\infty}S_{2n}=L\leq a_{1}$
and
(6)\displaystyle \begin{aligned} \lim_{n\rightarrow\infty}S_{2n+1} &=\lim_{n\rightarrow\infty}(S_{2n}+a_{2n+1})\\ &=\left(\lim_{n\rightarrow\infty}S_{2n}\right)+\left(\lim_{n\rightarrow\infty}a_{2n+1}\right)\\ &=L+0=L. \end{aligned}
Conclusion: ${\displaystyle\sum^{\infty}_{n=1}(-1)^{n+1}a_{n}=L}$.

Example 1. Consider the series
(7)$\displaystyle \sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n}$
We see that ${1/n\geq1/(n+1)}$ for each ${n}$, and
(8)$\displaystyle \lim_{n\rightarrow\infty}\frac{1}{n}=0.$
Thus the Alternating Series Test implies the alternating Harmonic series converges.

Example 2. Consider the series
(9)$\displaystyle \sum^{\infty}_{n=1}\frac{(-1)^{n+1}n}{(n+1)(n+2)}$
We see
(10)$\displaystyle \frac{n}{(n+1)(n+2)}\geq\frac{n+1}{(n+2)(n+3)};$
why? Well, multiply both sides by ${(n+2)}$ and we get
(11)$\displaystyle \frac{n}{n+1}\geq \frac{n+1}{n+3}$
Cross multiplication gives us
(12)$\displaystyle n(n+3)\geq (n+1)^{2} \iff n^{2}+3n \geq n^{2}+2n+1.$
Subtracting ${n^{2}+2n}$ from both sides gives us
(13)$\displaystyle n\geq1.$
So our series satisfies the first condition for the alternating series test.

We also see that
(14)$\displaystyle \frac{n}{(n+1)(n+2)}\approx\frac{1}{n}$
for “large ${n}$”. So we see
(15)$\displaystyle \lim_{n\rightarrow\infty}\frac{n}{(n+1)(n+2)}=0.$
Thus our series satisfies the criteria for the alternating series test, which implies convergence.

2. Definitions. A series ${\sum a_{n}}$ is “Absolutely Convergent” (or “Converges Absolutely”) if ${\sum|a_{n}|}$ converges.

On the other hand, if ${\sum|a_{n}|}$ is divergent, then we call the series ${\sum a_{n}}$ “Conditionally Convergent” (or we say the series “Converges Conditionally”).

Example 3. We see that ${\sum(-1)^{n}/n}$ is conditionally convergent, since ${\sum 1/n}$ diverges.

Example 4. The series ${\sum (-1)^{n}n^{-3/2}}$ is absolutely convergent since the integral test tells us ${\sum n^{-3/2}}$ converges.

Question: let ${\sum a_{n}}$ be a convergent series, and ${a_{n}\geq0}$ for each ${n}$. Does the series ${\sum (-1)^{n}a_{n}}$ converge?

Stop and think before continuing!

3. Absolute Convergence Test. If ${\sum |a_{n}|}$ conveges, then ${\sum a_{n}}$ converges.

Proof: We see first that
(16)$\displaystyle -|a_{n}|\leq a_{n}\leq|a_{n}|\quad\mbox{for each }n$
So what? Well, we see that
(17)$\displaystyle 0\leq a_{n}+|a_{n}|\leq2|a_{n}|\quad\mbox{for each }n$
Since ${\sum|a_{n}|}$ converges, we see ${\sum2|a_{n}|}$ converges too. But by the comparison test, we see
(18)$\displaystyle \sum^{\infty}_{n=1}a_{n}+|a_{n}|$
converges. So what? We haven’t proven ${\sum a_{n}}$ converges, have we? Consider the following trick
(19)$\displaystyle \sum a_{n} = \underbrace{\sum a_{n}+|a_{n}|}_{\text{converges}}-\underbrace{\sum |a_{n}|}_{\text{converges}}.$
Therefore the series ${\sum a_{n}}$ converges.

Example 5. Consider the series
(20)$\displaystyle \sum^{\infty}_{n=1}\frac{\cos(n)}{n^{3/2}}.$
What to do? We know
(21)$\displaystyle \left|\frac{\cos(n)}{n^{3/2}}\right|\leq\frac{1}{n^{3/2}}.$
So what? We know ${\sum n^{-3/2}}$ converges by the integral test. Then
(22)$\displaystyle \sum^{\infty}_{n=1}\left|\frac{\cos(n)}{n^{3/2}}\right|$
converges by comparison. By the absolute convergence test, we know our series in Equation (20) converges.

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## About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
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