What is $\lim_{n \rightarrow \infty}\sum\limits_{k = 1}^n\frac{k}{n^2}$? We have $$\dfrac{1+2+3+...+ \space n}{n^2}$$
What is the limit of this function as $n \rightarrow \infty$?
My idea:
$$\dfrac{1+2+3+...+ \space n}{n^2} = \dfrac{1}{n^2} + \dfrac{2}{n^2} + ... + \dfrac{n}{n^2} = 0$$
Is this correct?
 A: HINT:
Such application of limit (esp. to $\infty$) to individual summand is applicable only when the number of summand is finite.
$$\dfrac{1+2+3+...+ \space n}{n^2} =\frac{\frac{n(n+1)}2}{n^2}=\frac12\cdot\left(1+\frac1n\right)$$
A: Another method, using Riemann sums applied to $f:x\mapsto x$:
$$
\frac{\sum_{k=1}^n k }{n^2} = \frac{1}{n}\sum_{k=1}^n \frac{k }{n}  \xrightarrow[n\to\infty]{}
\int_{0}^{1} xdx
$$
As for your question, each term goes to $0$, but the number of terms grows; you are not in the case where you have the sum of "constantly many terms, each of them converging".
A: Recall that $$1 + 2 + \cdots + n =\sum_{i = 1}^n i = \frac{n(n+1)}{2}$$
So you want $$\lim_{n\to \infty}\dfrac{n(n+1)}{2n^2} = \lim_{n\to\infty}\frac{n^2+ n}{2n^2} = \lim_{n\to \infty} \left(\frac 12 + \frac 1{2n}\right)$$
A: Your method is wrong. Consider the first values of the expression:


*

*for $n=1$ you get $1$;

*for $n=2$ you get $1/4+2/4=3/4$;

*for $n=3$ yoy get $1/9+2/9+3/9=6/9=2/3$.


One could make the conjecture that these terms are always bigger than $1/2$, which can be proved by induction. Let
$$
f(n)=\frac{1}{n^2}\sum_{i=1}^n i
$$
We have $f(1)=1>1/2$. Suppose the assertion holds for $f(n)$ and consider
\begin{align}
f(n+1)&=\frac{1}{(n+1)^2}\sum_{i=1}^{n+1} i\\
&=\biggl(\frac{1}{(n+1)^2}\sum_{i=1}^{n} i\biggr)+\frac{n+1}{(n+1)^2}\\
&=\frac{n^2}{(n+1)^2}\biggl(\frac{1}{n^2}\sum_{i=1}^n i\biggr)+\frac{1}{n+1}\\
\text{(by induction hypothesis)}\qquad&>\frac{1}{2}\frac{n^2}{(n+1)^2}+\frac{1}{n+1}\\
&=\frac{1}{2}\frac{n^2+2n+2}{(n+1)^2}\\
&=\frac{1}{2}\biggl(\frac{n^2+2n+1}{(n+1)^2}+\frac{1}{(n+1)^2}\biggr)\\
&=\frac{1}{2}\biggl(1+\frac{1}{(n+1)^2}\biggr)\\
&>\frac{1}{2}
\end{align}
Therefore your conclusion that the limit is $0$ cannot be true.
