Is $\sum\limits_{k=1}^{n}\frac{1}{(n+k)^2}$ divergent or convergent?

Question

Problem: In a sequence $\{a_n\}_{n=1}^\infty,$ $$a_n=\sum_{k=1}^{n}\frac{1}{(n+k)^2}.$$

Determine if the sequence is convergent or divergent and, if convergent, compute its limit.

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Honestly, I don't really know where to start. In my book they state the following regarding convergence for sequences:

The sequence $\{a_n\}_{n=1}^\infty$ is said to converge with limit $a$ if $\forall \ \varepsilon >0 \ \exists \ N_\varepsilon:n>N_\varepsilon\Rightarrow|a_n-a|<\varepsilon.$

What does this even mean? I know the symbols but I can't intuitively understand it and apply it on my problem.

Answer 1

It is convergent and the limit is $0$, because:

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

and both the sequence $0$ and $\frac {1}{n}$ converge to $0$.

Answer 2

$$a_n= \sum_{k=1}^{n} \frac{1}{(n+k)^2} =\frac{1}{n}\left(\frac{1}{n}\sum_{k=1}^{n} \frac{1}{(1+\frac{k}{n})^2}\right) \to0$$ since by Riemann sum. $$\sum_{k=1}^{n} \frac{1}{(n+k)^2} =\frac{1}{n}\sum_{k=1}^{n} \frac{1}{(1+\frac{k}{n})^2} \to\int_0^1 \frac{dx}{(1+x)^2}$$

Answer 3

Note that $$ \frac{1}{(n+k)^2}\leq \frac{1}{n^2} $$ so $$ 0\leq a_n=\sum_{k=1}^{n}\frac{1}{(n+k)^2}\leq n\times \frac{1}{n^2}\to0 $$ Alternatively $$ \sum_{k=1}^{\infty}\frac{1}{k^2}<\infty $$ by the integral test or cauchy condensation test. Hence $$ 0\leq a_n=\sum_{k=1}^{n}\frac{1}{(n+k)^2}=\sum_{k=n+1}^{2n}\frac{1}{k^2}\to0 $$ since the partial sums are cauchy.

Answer 4

Note that:$$\sum_{k=1}^{n}\frac{1}{(n+k)^2}=\sum_{k=1}^{2n}\frac{1}{k^2}-\sum_{k=1}^{n}\frac{1}{k^2}$$