What does the following sequence converge to?

Question

Does the following sequence converge or diverge? If it converges, find its limit.

$\frac{1^1}{n^2}+\frac{2^1}{n^2}+........+\frac{n^1}{n^2}$

I tried to get a general formula for this sequence and simply got:

$\frac{n}{n^2}$ or $\frac{1}{n}$

This led me to think that the sequence converges to $0$ since the limit of this function is $0$. But this limit is incorrect. This tells me the general formula I got for the sequence is incorrect. But I am not sure why.

Any help?

Answer 1

You want to calculate the following $$\sum_{k=1}^n \frac{k}{n^2}$$ It is equal to $$\dfrac{\sum_{k=1}^n k}{n^2}$$ $$\sum_{k=1}^n k = {n(n+1)\over 2}$$ $$\dfrac{\sum_{k=1}^n k}{n^2} = \dfrac{n(n+1)}{2n^2}= \dfrac{(n+1)}{2n}$$ $$\lim_{n\to \infty} \dfrac{(n+1)}{2n}=\frac{1}{2}$$

Answer 2

Note that $$\sum_{k=1}^n k = {n(n+1)\over 2}.$$

Answer 3

We have that

$$\frac{1}{n^2}+\frac{2}{n^2}+........+\frac{n}{n^2}=\frac1{n^2} \sum_{k=1}^n k$$

then recall that $\sum_{k=1}^n k = {n(n+1)\over 2}$.

More in general for

$$\frac{1^p}{n^{p+1}}+\frac{2^p}{n^{p+1}}+........+\frac{n^p}{n^{p+1}}=\frac1{n^{p+1}} \sum_{k=1}^n k^p$$

and by Faulhaber's formula

$$\sum_{k=1}^nk^p \sim \frac{n^{p+1}}{p+1}$$

Answer 4

Your expression is a Riemann sum. To be more precise, it is the upper sum of the function $f(x)=x$ defined on $[0,1]$ with respect to the partition $\left\{0,\frac1n,\frac2n,\ldots,\frac nn(=1)\right\}$. Therefore$$\lim_{n\to\infty}\frac1{n^2}+\frac2{n^2}+\cdots+\frac n{n^2}=\int_0^1x\,\mathrm dx=\frac12.$$

Answer 5

By Stolz Cesàro,

$$\lim_{n \to \infty} \frac{1+2+3+...+ n}{n^2}$$

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

$$=\frac{1}{2}$$