Other Alternating Series Tests

1. We have another couple of methods testing if an alternating series converges. It’s worth knowing as many different ways as possible, because sometimes one doesn’t work well (or at all).

We will consider a couple tests. For each test we provide a proof that it works, and a couple examples.


2. Alternating Ratio Test. Consider the series
(1)\displaystyle  \sum^{\infty}_{n=0}(-1)^{n}a_{n}
Let
(2)\displaystyle  L = \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_{n}}.

  1. If {L<1}, then the series {\sum (-1)^{n}a_{n}} conveges absolutely.
  2. If {L>1} (or {L=\infty}), then the series {\sum (-1)^{n}a_{n}} diverges.

Proof of Convergence: We will prove if {L<1}, then the series
(3)\displaystyle  \sum^{\infty}_{n=0}a_{n}
converges. The absolute convergence test implies the alternating series will converge. What to do?

We first consider some {\varepsilon} satisfying {L<\varepsilon<1}. (Can we do this? Sure, pick {(L+1)/2}, and we’re good!) Since we suppose {L<1}, then there exists an {N} such that
(4)\displaystyle  \left|\frac{a_{n+1}}{a_{n}}\right|<r\quad\mbox{for any }n\geq N.
We have
(5)\displaystyle  \begin{aligned} a_{N+1}&<r a_{N}\\ a_{N+2}&<r a_{N+1}<r^{2}a_{N}\\ a_{N+3}&<r^{3}a_{N}\\ a_{N+k}&<r^{k}a_{N} \end{aligned}
So we form a geometric series
(6)\displaystyle  \sum^{\infty}_{k=0}a_{N}r^{k} = \frac{a_{N}}{1-r}
which bounds the “most” of our series
(7)\displaystyle  0\leq\sum^{\infty}_{k=0}a_{N+k}\leq\sum^{\infty}_{k=0}a_{N}r^{k}
The comparison test tells us that “most” of our series converges. But what about our whole series? We write it as
(8)\displaystyle  \sum^{\infty}_{n=0}a_{n} = \underbrace{\sum^{N-1}_{n=0}a_{n}}_{\text{finite}}+\underbrace{\sum^{\infty}_{k=0}a_{N+k}}_{\text{converges}}
so it converges.

Proof of Divergence: We assume that {L>1}. The ratio {a_{n+1}/a_{n}} will eventually be greater than 1, too. So there exists an {N} such that
(9)\displaystyle  \left|\frac{a_{n+1}}{a_{n}}\right|>1\quad\mbox{for any }n\geq N.
So we see {|a_{n+1}|>|a_{n}|} whenever {n\geq N}, thus
(10)\displaystyle  \lim_{n\rightarrow\infty}a_{n}\not=0.
Thus it’s impossible for the series {\sum a_{n}} to converge!

Remark 1. We only really proved that the series doesn’t converge absolutely. If we pick some {r} between {1<r<L}, then there exists an {N} such that
(11)\displaystyle  \left|\frac{a_{n+1}}{a_{n}}\right|>r\quad\mbox{for any }n\geq N.
So we have
(12)\displaystyle  a_{N+k} > r^{k}a_{N}.
We have our series
(13)\displaystyle  \begin{aligned} \sum^{\infty}_{k=1}(-r)^{k}a_{N} &= a_{N} \sum^{\infty}_{k=1}(r^{2k}-r^{2k-1})\\ &= a_{N} (r-1)\sum^{\infty}_{k=1}r^{2k}. \end{aligned}
We see since {r>1} that the series
(14)\displaystyle  \sum^{\infty}_{k=1}r^{2k}\quad\mbox{diverges}
Thus
(15)\displaystyle  \sum^{\infty}_{k=1}(-r)^{k}a_{N}\quad\mbox{diverges}
We see then that the series
(16)\displaystyle  \sum^{\infty}_{k=1}a_{N+k}\geq\sum^{\infty}_{k=1}(-r)^{k}a_{N}
diverges by the comparison test.

3. The Root Test. Consider the series
(17)\displaystyle  \sum^{\infty}_{n=1}(-1)^{n}a_{n}
Let
(18)\displaystyle  L = \lim_{n\rightarrow\infty}\sqrt[n]{a_{n}}

  1. If {L<1}, then the series converges absolutely;
  2. If {L>1} (or {L=\infty}), then the series diverges.

Proof: It’s similar to the ratio test, for the convergent case we pick some {r} satisfying {L<r<1}. Then we have some {N} satisfying
(19)\displaystyle  \sqrt[n]{a_{n}}<r\quad\mbox{for any }n\geq N.
which implies
(20)\displaystyle  a_{n}<r^{n}.
Thus we have
(21)\displaystyle  \sum^{\infty}_{k=1}a_{N+k}<\sum^{\infty}_{k=1}r^{N+k}<\sum^{\infty}_{k=0}r^{k}=\frac{1}{1-r}
which by the comparison test implies convergence.
The proof for divergence is similar.

Remark 2. When {L=1}, the ratio test doesn’t say anything about convergence or divergence for the series.

Exercise 1. Determine the convergence or divergence of \displaystyle\sum^{\infty}_{n=1}\frac{(-1)^{n}}{n\cdot\ln(n+2)}.

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About Alex Nelson

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

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