Convergence of $\sum_{n=1}^{\infty} \frac{1}{n^2}$ by comparing with $\frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \ldots+$

Question

I am trying to prove that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges using the series $\sum_{n=1}^{\infty} \frac{1}{2^{n-1}}$.

I know that the latter series converges to 2. Each term of the latter series is bigger than the corresponding term in the former series so the former series converges. I think the proof right now is already sufficient.

However, in my handwritten notes I see something that says "Twice of $\sum_{n=1}^{\infty} \frac{1}{2^{n-1}}$ converges, so $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges. Why do we need to look at twice of $\sum_{n=1}^{\infty} \frac{1}{2^{n-1}}$, i.e., $\sum_{n=1}^{\infty} \frac{1}{2^{n-2}}$, instead of just $\sum_{n=1}^{\infty} \frac{1}{2^{n-1}}$?

I could have miswritten something, so I just want to make sure the looking at $\sum_{n=1}^{\infty} \frac{1}{2^{n-1}}$ alone is sufficient.

Answer 1

I like this one...

For $n> 1$

$\frac{1}{n^2} < \frac {1}{n(n-1)}$

$\sum_\limits{n=1}^\infty \frac{1}{n^2} < 1 + \sum_\limits{n=2}^\infty \frac{1}{n(n-1)} = 1 + \sum_\limits{n=2}^\infty (\frac{1}{n-1} - \frac{1}{n})$

$\sum_\limits{n=2}^N (\frac{1}{n-1} - \frac{1}{n}) = 1-\frac {1}{N}$

Taking the limit as N approaches infinity...

$\sum_\limits{n=1}^\infty \frac{1}{n^2} < 2$

What it looks like you have been asked for, though, is to break up this series into blocks like so $(1),(2,3), (4,5,6,7), \cdots.$ With succeeding block twice as big as the one before...

And for each block...
$\sum_\limits {n=2^k}^{2^{k+1}-1}\frac 1{n^2} < 2^k\frac {1}{2^{2k}} = \frac {1}{2^k}$

$\sum_\limits{n=1}^\infty \frac{1}{n^2} < \sum_\limits{n=1}^\infty \frac 1{2^n}$

This can be generalized to show that the series $\sum_\limits{n=1}^\infty \frac{1}{n^p}$ converges for all $p>1$

$\sum_\limits{n=1}^\infty \frac{1}{n^p} < \sum_\limits{n=1}^\infty 2^{(1-p)n}$

Answer 2

We can use the idea of Cauchy condensation test which leads to

$$\sum_{n=1}^{\infty}\frac1{n^2} \le \sum_{n=0}^{\infty}\frac{2^n}{2^{2n}}=\sum_{n=0}^{\infty}\frac{1}{2^{n}}=\sum_{n=1}^{\infty}\frac{1}{2^{n-1}} $$

that is

$$1+\frac14+\frac19+\frac1{16}+\ldots+\frac1{n^2}+\ldots \le 1+2\left(\frac1{4} \right)+2^2\left(\frac1{16}\right)+2^3\left(\frac1{64}\right)+\ldots$$

therefore the factor $2$ is not necessary at all even of course it leads to a sufficient condition for the convergence.