Evaluating $\lim_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots+\frac{n}{n^2+n}\right)$.

Question

$$\lim_{n\to \infty}\left[\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right]$$

For calculating the above limit, why can't we distribute limit over each term ?

If we distribute the limit over each term then by observing dominating term of numerator and denominator of each, limit will come out to be 0 for each expression thus final answer will be $0$. But this is not the case and answer is 1 also here we can't distribute the limit inside ?? Why??

My resource solves this question using Sandwich theorem. First smallest term is calculated then largest term, now all terms in the limit lie in between them (equality also). So using sandwich theorem limit comes out to be $1$.

Please explain.

Answer 1

You can't distribute the limit since the number of terms is dependent on $n$. Another way of expressing the limit is as $$\lim_{n\to \infty}\sum_{i=1}^n\frac{n}{n^2+i}$$ By distributing, you'd be doing $$\sum_{i=1}^n\lim_{n\to \infty}\frac{n}{n^2+i}$$ Which neglects the $n$ in the index of the sum.

Notice that $$a_n:=\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}$$ has $n$ terms all greater than $\frac{n}{n^2+n}$ so $$\forall n\in\mathbb{N},\; \frac{n^2}{n^2+n}\leq a_n$$ and thus $\lim a_n\neq0$.

Answer 2

Your limit isn't equal to the sum of the limits, because the theorem "the limit of sum equals the sum of limits" is valid for the sum of a finite amount of terms; in this case $n \to \infty$, so the terms you are summing are infinite.

Hint: the denominators are always bigger than $n^2+1$ and always lesser than $n^2+n$.

Answer 3

We have $${1\over n}\ge {n\over n^2+i}\ge {1\over n+1}$$ Hence the limit of the sum is equal $1.$