Series 1 - 1/2^2 + 1/3 - 1/4^2 + 1/5 - 1/6^2 + 1/7...

Question

This is the exercise 2.7.2 e) of the book "Understanding Analysis 2nd edition" from Stephen Abbott, and asks to decide wether this series converges or diverges: $$ 1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}-\dfrac{1}{8^2}\dots $$

I have noticed that $$\dfrac{1}{3}< 1-\dfrac{1}{2^2}+\dfrac{1}{3};\\ \dfrac{1}{3}+\dfrac{1}{5}<1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5};\\ \dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}<1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}\\ $$ which is true in general because $$1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{36}-\dfrac{1}{64}-\dots=1-\sum_{n=1}^\infty \dfrac{1}{(2n)^2}=1-\dfrac{1}{4}\sum_{n=1}^\infty\dfrac{1}{n^2}=1-\dfrac{\pi^2}{24}>0. $$ Thus $$ \sum_{n=1}^\infty \dfrac{1}{2n+1}<1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}-\dfrac{1}{8^2}\dots $$ Finally, as the series $\sum_{n=1}^\infty \dfrac{1}{2n+1}$ diverges, so does the series requested.

My two questions are:

I) Is this reasoning correct?

II) Can this exercise be done without using that $\sum_{n=1}^\infty\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$? I would like to find a solution with more elementary tools.

Thank you.

Answer 1

The basic idea is good. But after you've proven that $\sum_{n=1}^N\frac1{2n+1}$ is smaller that the sum of the first $2N+1$ terms of your series, you're done. And all you need for that is that$$\frac1{2^2}+\frac1{4^2}+\cdots+\frac1{(2N)^2}<1.$$This is equivalent to$$\frac1{1^2}+\frac1{2^2}+\cdots+\frac1{N^2}<4,$$which in turn follows from\begin{align}\frac1{1^2}+\frac1{2^2}+\cdots+\frac1{N^2}&<1+\frac1{1\times2}+\frac1{2\times3}+\cdots+\frac1{(N-1)N}\\&=1+1-\frac12+\frac12-\frac13+\cdots+\frac1{N-1}+\frac1N\\&=2+\frac1N\\&<4.\end{align}

Answer 2

Don't be swayed by the particular case. You have the following

Lemma Let $\sum a_n$ be a series (resp. $\sum b_n$ be a convergent series), then the intertwining $x_{2n}=a_n$ and $x_{2n+1}=b_n$ is convergent iff $\sum a_n$ is so.

Proof Let us, for short, suppose the indexing starting from zero. Then, we get
$$ \sum_{n=0}^{2N+1}x_n=\sum_{n=0}^N a_n + \sum_{n=0}^N b_n $$ this proves the equivalence $$ \sum x_n\mbox{ converges }\Longleftrightarrow \sum a_n\mbox{ converges } $$ end of proof

Here $$ 1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}\ldots $$ diverges while $$\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}\ldots $$ converges. Hence your series diverges.

Answer 3

$${1\over2n-1}-{1\over(2n)^2}={4n^2-2n+1\over4n^2(2n-1)}\gt{4n^2-4n+1\over4n^2(2n-1)}={2n-1\over4n^2}\ge{1\over4n}$$

so

$$\left(1-{1\over4}\right)+\left({1\over3}-{1\over16}\right)+\left({1\over5}-{1\over36}\right)+\cdots\gt{1\over4}+{1\over8}+{1\over12}+\cdots={1\over4}\left(1+{1\over2}+{1\over3}+\cdots\right)$$