find limit of $a_n=\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+...+\frac{1}{(2n)^2}$

Question

finding limit of $a_n=\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+...+\frac{1}{(2n)^2}$

I know that i have to use Stolz Cesaro theorem, but the problem is that i need second sequence.

Answer 1

The limit is $0$:

$\displaystyle\frac1{(n+k)^2}\le\frac1{n^2}$ for $1\le k\le n$, so $\displaystyle 0\le a_n\le \frac{n}{n^2}=\frac1n\to0$.

Let me add two remarks: First of all, $\sum_n a_n$ diverges, because if fact $a_n$ grows like $1/n$: $\displaystyle\frac1{(n+k)^2}\ge\frac1{(2n)^2}$ for $1\le k\le n$, so $\displaystyle a_n\ge \frac{n}{(2n)^2}=\frac1{4n}$.

Second, the above suggests to study $\lim_{n\to\infty}n a_n$. Recognizing this as a Riemann sum is the key: Note that $$ n\sum_{k=1}^n\frac1{(n+k)^2}=\frac1n\sum_{k=1}^n\frac1{(1+k/n)^2}, $$ which we should recognize as a Riemann sum for $\displaystyle \int_0^1\frac1{(1+x)^2}\,dx$, so the two expressions coincide when $n\to\infty$, and we see that $$\lim_{n\to\infty}na_n=\int_0^1\frac1{(1+x)^2}\,dx=\left.\frac{-1}{1+x}\right|_0^1=\frac12. $$ To illustrate the rate of convergence, Sage evaluates $na_n$ when $n=200$, as $0.498128645813151$; when $n=2000$, as $0.499812536458331$; and when $n=20000$, as $0.499981250364583$.

Answer 2

HINT:

$$S=\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+...+\frac{1}{(2n)^2}=\sum_{1\le r\le 2n}\frac1{r^2}-\sum_{1\le r\le n}\frac1{r^2}$$

Putting $2n=m$ in the first sum $$S=4\sum_{1\le r\le m}\frac1{r^2}-\sum_{1\le r\le n}\frac1{r^2}$$

Now use this

Answer 3

$$a_n=\frac{1}{n^2}\left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right)=\frac{1}{n}\left[\frac{1}{n}\left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right)\right]$$

Now, apply S-C to

$$a_n=\frac{ \left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right) }{n^2}$$

But this is an overkill, it is easy to see that the top is $\leq 1+1+1..+1=n$.

P.S.

$$\frac{1}{n}\left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right)$$ is a Riemann sum, thus convergent, and $\frac{1}{n}$ converges to $0$.

Alternately, if you don't know integrals,apply S-C to

$$a_n=\frac{ \left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right) }{n^2}$$

Answer 4

You can solve by another method . This must be interesting to you.

Hint : $lt_{n \rightarrow \infty} $ $\frac{1}{n} \sum_{k=1}^{n} f(\frac{k}{n})$ $ = $ $\int_0^1 f(x) dx$ where $f$ is in $C[0,1]$