Prove the sequence $\sum_{k=0}^n \frac{1}{(n+k)^2}$ is convergent

Question

I'm having a hard time trying to prove that the sequence $\{a_n\}$ whose general term $a_n$ is

$$\sum_{k=0}^n \frac{1}{(n+k)^2}$$

is convergent. I'm trying to prove it by definition, that is to say, by finding a lower/upper bound and by proving that it is decreasing/increasing using induction.

By subtracting $a_n$ from $a_{n+1}$ we obtain the following for $n=m$ (if I'm not mistaken):

$$ \frac{1}{(2m+1)^2} + \frac{1}{(2m+2)^2} - \frac{1}{m^2} $$

Which is less than $0$ for $n=1$ from which I have assumed the sequence is decreasing and therefore trying to prove that $a_{n+1} - a_n < 0$ using induction. But I am terribly stuck! Thanks in advance for any suggestion.

Answer 1

To prove that this sequence converges, we will use the monotone convergence theorem which states

If $(a_n)$ is either

1. increasing and bounded above
2. decreasing and bounded below

then it is a convergent sequence.

In order to use this result, we first look at the difference $a_{n+1}-a_{n}$ to determine if your sequence is increasing or decreasing. We have \begin{align} a_{n+1}-a_n &= \sum_{k=0}^{n+1} \frac{1}{(n+1+k)^2} - \sum_{k=0}^n \frac{1}{(n+k)^2}\\ &=\sum_{k=0}^{n+1} \frac{1}{(n+(k+1))^2} - \sum_{k=0}^n \frac{1}{(n+k)^2}\\ &=\sum_{k=1}^{n+2} \frac{1}{(n+k)^2} - \sum_{k=0}^n \frac{1}{(n+k)^2}\\ &=\left(\sum_{k=1}^{n} \frac{1}{(n+k)^2} \frac{1}{(2n+1)^2} + \frac{1}{(2n+2)^2} +\right) - \left(\frac{1}{n^2}+\sum_{k=1}^n \frac{1}{(n+k)^2}\right)\\ &=\frac{1}{(2n+1)^2} + \frac{1}{(2n+2)^2} - \frac{1}{n^2}\\ &\leq \frac{1}{(2n)^2} + \frac{1}{(2n)^2} -\frac{1}{n^2}\\ &= -\frac{1}{2n^2}\leq 0 \end{align} So your sequence is decreasing. Furthermore, it is clear that $a_n\geq 0$ for any $n\in\mathbb{N}$. That is, your sequence is bounded below. By the monotone convergence theorem, it converges.

Answer 2

You don't need induction. $\frac{1}{(2m+2)^2}+\frac{1}{(2m+1)^2}-\frac{1}{m^2}\lt \frac{2}{(2m)^2}-\frac{1}{m^2}=\frac{1}{2m^2}-\frac{1}{m^2}\lt 0$.

Answer 3

It might be easier to bound the sum from above by a telescoping sum and then invoke the Squeeze Theorem:

$$0\le\sum_{k=0}^n{1\over(n+k)^2}\lt\sum_{k=0}^n{1\over(n+k-1)(n+k)}=\sum_{k=0}^n\left({1\over n+k-1}-{1\over n+k}\right)={1\over n-1}-{1\over2n}={n+1\over2n(n-1)}\to0$$

Answer 4

Hint:

\begin{align*} \lim_{n\to\infty}\sum_{k=1}^n\frac{1}{(n+k)^2}\leq\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k^2} \end{align*}

We also have $\sum_{k=1}^\infty\frac{1}{k^\alpha}$ is convergent iff $\alpha>1$.

Answer 5

Just for your curiosity since you already received good answers.

$$S_n=\sum_{k=0}^n \frac{1}{(n+k)^2}=\psi ^{(1)}(n)-\psi ^{(1)}(2 n+1)$$ where appears the polygamma function (don't worry : you will learn about it sooner or later).

Using the asymptotics for large values of $n$ $$S_n=\frac{1}{2 n}+\frac{5}{8 n^2}+O\left(\frac{1}{n^3}\right)$$ For example $$S_{10}=\frac{3056206830982561}{54192375991353600}\approx 0.0563955$$ while the above expansion gives $\frac{9}{160}=0.05625$

Answer 6

$\displaystyle\sum_{k = 0}^{n}{1 \over \left(n + k\right)^{2}} = {1 \over n}\left[{1 \over n}\sum_{k = 0}^{n}{1 \over \left(1 + k/n\right)^{2}}\right] \,\,\,\stackrel{\color{red}{\mathrm{as}\ n\ \to\ \infty}}{\sim}\,\,\, {1 \over n}\int_{0}^{1}{\mathrm{d}x \over \left(1 + x\right)^{2}} = {1 \over 2n}$