Limit of the sequence $\frac{1}{n^2+1}+\frac{2}{n^2+2}+...+\frac{n}{n^2+n}$

Question

As in the title, we have to calculate the limit $$\frac{1}{n^2+1}+\frac{2}{n^2+2}+...+\frac{n}{n^2+n}$$ when $n\rightarrow+\infty$. It is easy to see that $$\lim_{n\to\infty} \frac{1}{n^2+1}+\frac{1}{n^2+2}+...+\frac{1}{n^2+n}=0$$ and $$\lim_{n\to\infty} \frac{n}{n^2+1}+\frac{n}{n^2+2}+...+\frac{n}{n^2+n}=1$$ I don't know how to calculate the limit. I am pretty sure that we somehow apply the sandwich rule, but I am afraid that there is something similary with $$\lim_{n\to\infty} \frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2}=\frac{\pi^2}{6}$$

Answer 1

Replace all the denominators by $n^2+n$, and then by $n^2+1$, to use squeeze theorem and obtain the answer to be $\frac 12$.

Answer 2

Hint: $$\lim_{n\to\infty} \frac{1}{n^2+1}+\frac{2}{n^2+1}+...+\frac{n}{n^2+1}=\frac{1}{2}$$ and

$$\lim_{n\to\infty} \frac{1}{n^2+n}+\frac{2}{n^2+n}+...+\frac{n}{n^2+n}=\frac{1}{2}$$

Answer 3

Convince yourself that $$\frac1{n^2+n}+\frac2{n^2+n}+\cdots+\frac n{n^2+n}\le \frac1{n^2+1}+\frac2{n^2+2}+\cdots+\frac n{n^2+n} \le$$ $$\le \frac1{n^2}+\frac2{n^2}+\cdots+\frac n{n^2},$$ and remember that $1+2+\cdots+n=\tfrac {n(n+1)}2$. Can you apply the Sandwich rule now?

Answer 4

We can rewrite your sum in terms of harmonic numbers and then apply the well-known asymptotics of the latter.

Here we go

$$s = \frac{1}{1+n^2}+ \frac{2}{2+n^2}+ \text{...}+\frac{n}{n+n^2}$$

$$ = (\frac{1}{1+n^2}-1)+ (\frac{2}{2+n^2}-1)+ \text{...}+(\frac{n}{n+n^2}-1) + n*1$$

$$ = n+ (\frac{-n^2}{1+n^2})+ (\frac{2-n^2+2}{2+n^2})+ \text{...}+(\frac{n-n^2-n}{n+n^2}) $$

$$ = n - n^2\left( (\frac{1}{1+n^2})+ (\frac{1}{2+n^2})+ \text{...}+(\frac{1}{n+n^2}) \right)$$

$$ = n-n^2 \sum_{k=1+n^2}^{n+n^2} 1/k$$

This can be written in terms of the harmonic number $H_n = \sum_{k=1}^n 1/k$ as

$$s = n-n^2(H_{n+n^2} - H_{n^2})$$

Now we can calculate the limit $n \to \infty$.

Since in the limit $x\to\infty$ we have

$$H_x \sim \log(x)+\gamma$$

we find

$$s \sim n - n^2(\log(n+n^2)-\log(n^2)) $$

$$s \sim n - n^2(\log(1+1/n)) $$

Using the expansion of the $\log$ this becomes finally

$$s \sim n - n^2(\frac{1}{n} - \frac{1}{2n^2}) = 1/2$$

as expected.

Answer 5

Another idea is to identify the sum

$$s = \sum _{k=1}^n \frac{k}{k+n^2}$$

in the limit $n\to \infty$ as a Riemann integral.

In fact we can write

$$s = \frac{1}{n} \sum _{k=1}^n \frac{k}{n \left(\frac{k}{n^2}+1\right)}$$

For $n\to\infty$ we can neglect $\frac{k}{n^2}$ against unity:

$$s = \frac{1}{n} \sum _{k=1}^n \frac{k}{n} $$

which asymptotically goes to

$$s \to \int_0^1 x \, dx = 1/2$$

An obvious generalization of the sum gives immediately

$$s_3 = \sum _{k=1}^n \frac{k^2}{k^2+n^3}\to \int_0^1 x^2 \, dx = 1/3 $$

and more generally

$$s_m = \sum _{k=1}^n \frac{k^{m-1}}{k^{m-1}+n^m}\to \int_0^1 x^{m-1} \, dx = 1/m $$

The latter relation holds even for any real $m \gt 0$.