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
(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}
(5)\displaystyle  \lim_{n\rightarrow\infty}S_{2n}=L\leq a_{1}
(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.


About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Analysis, Calculus, Infinite Series. Bookmark the permalink.

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