# Math Help - Prove that the series 1/n^2 converges

1. ## Prove that the series 1/n^2 converges

Prove that the series whose terms are 1/n^2 converges by showing that the partial sums form a Cauchy sequence.

I've tried to start this as follows: Assuming that m>n, we have

|a_n-a_m|=1/m^2+1/(m+1)^2+...+1/(n+1)^2 <= (m-n)/(n+1)^2.

So to show it's Cauchy, I need to find N such that m,n>N implies |a_n-a_m|<epsilon for any epsilon. But how can I find this? Don't I have to somehow eliminate m from the expression? I don't know how to do this. Help!

2. ## Series not sequence

Originally Posted by EMR
Prove that the series whose terms are 1/n^2 converges by showing that the partial sums form a Cauchy sequence.

I've tried to start this as follows: Assuming that m>n, we have

|a_n-a_m|=1/m^2+1/(m+1)^2+...+1/(n+1)^2 <= (m-n)/(n+1)^2.

So to show it's Cauchy, I need to find N such that m,n>N implies |a_n-a_m|<epsilon for any epsilon. But how can I find this? Don't I have to somehow eliminate m from the expression? I don't know how to do this. Help!

I think what you need to show is this is less than epsilon

$S_n=\sum_{i=1}^{n}\frac{1}{i^2}$ and $S_m=\sum_{i=1}^{m}\frac{1}{i^2}$

then
$|S_m-S_n|=|\sum_{i=1}^{m}\frac{1}{i^2}-\sum_{i=1}^{n}\frac{1}{i^2}|=|\sum_{i=n}^{m}\frac{ 1}{i^2}|$

See what you can do from here.

3. Originally Posted by EMR
Prove that the series whose terms are 1/n^2 converges by showing that the partial sums form a Cauchy sequence.

I've tried to start this as follows: Assuming that m>n, we have

|a_n-a_m|=1/m^2+1/(m+1)^2+...+1/(n+1)^2 <= (m-n)/(n+1)^2.

So to show it's Cauchy, I need to find N such that m,n>N implies |a_n-a_m|<epsilon for any epsilon. But how can I find this? Don't I have to somehow eliminate m from the expression? I don't know how to do this. Help!
You might be interested to know that the series converges to the value $\frac{\pi^2}{6}$. A (non-rigorous) proof was first given by Euler. It can be proved rigorously and easily using Fourier analysis.

4. Originally Posted by TheEmptySet
I think what you need to show is this is less than epsilon

$S_n=\sum_{i=1}^{n}\frac{1}{i^2}$ and $S_m=\sum_{i=1}^{m}\frac{1}{i^2}$

then
$|S_m-S_n|=|\sum_{i=1}^{m}\frac{1}{i^2}-\sum_{i=1}^{n}\frac{1}{i^2}|=|\sum_{i=n}^{m}\frac{ 1}{i^2}|$

See what you can do from here.
Yes, that's what I meant...couldn't figure out how to do the Latex code here.

Okay, so I need to show that
$|S_m-S_n||\sum_{i=1}^{m}\frac{1}{i^2}-\sum_{i=1}^{n}\frac{1}{i^2}|=|\sum_{i=n+1}^{m}\fra c{1}{i^2}|\le \frac{m-n}{(n+1)^2}<\frac{m}{n^2}<\epsilon$.

But again I am faced with finding an N such that m,n>N implies that this is less than any given epsilon. But to do that, I should somehow eliminate m, yes?

Thanks again!

5. You have to show that $\sum_{k=m+1}^n \frac{1}{k^2} < \epsilon$ for $n>m$. Note, $\frac{1}{k^2} \leq \frac{1}{(m+1)^2}$ for $m+1\leq k \leq n$. Thus, $\sum_{k=m+1}^n \frac{1}{k^2} \leq \sum_{k=m+1}^n \frac{1}{(m+1)^2} = \frac{n-m}{(m+1)^2} \leq \frac{n}{m^2}$. The problem is this is a bad estimate, you cannot make this quanity sufficiently small.

6. So is it even possible to directly show that the sequence of partial sums form a Cauchy sequence directly? Clearly they do, because the series itself converges, but I can't show it directly from the definition.

7. Originally Posted by EMR
So is it even possible to directly show that the sequence of partial sums form a Cauchy sequence directly? Clearly they do, because the series itself converges, but I can't show it directly from the definition.
It is possible to prove this using an integral estimate.

Consider the function, $f(x) = \frac{1}{x^2}$ defined on the positive reals. We will place a good bound on the quantity $\sum_{k=m+1}^n \frac{1}{k^2}$ where $n>m\geq 1$. Consider the function $f(x)$ on the interval $[m,n]$. The area is equal to $\int_m^n \frac{dx}{x^2} = \frac{1}{m} - \frac{1}{n}$. Now consider the right-endpoint rectangular approximation of the area equally spaced at integer points between $m$ and $n$ (each sub-rectangle has width $1$) this approximation is clearly smaller then the area, furthermore its quantity is $\sum_{k=m+1}^n \frac{1}{k^2}$.

Thus, $\sum_{k=m+1}^n \frac{1}{k^2} \leq \frac{1}{m} - \frac{1}{n} \leq \frac{1}{m}$. Note, $1/m$ can be made arbitrary small if we pick $N$ so large that $n>m>N$.

8. $\sum_{k=n}^m \frac{1}{k^2} < \sum_{k=n}^m \frac{1}{k(k-1)} = \sum_{k=n}^m \left( \frac{1}{k-1} - \frac{1}{k} \right) = \frac{1}{n-1} - \frac{1}{m} < \frac{1}{n-1}$