$$\lim_{n\to\ \infty}\frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}$$ For using Sandwich theorem I need two functions such that $g(x)<f(x)<h(x)$ $$\frac{1}{n+n^2} +\frac{2}{n+n^2}+\cdots+\frac{n}{n+n^2} \leq \frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2} $$ But I can't find function greater than the given which will help me evaluate the limit.
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$\begingroup$ See also: Evaluate the limit $\lim\limits_{n \to \infty} \frac{1}{1+n^2} +\frac{2}{2+n^2}+ \ldots +\frac{n}{n+n^2}$. Found using Approach0. $\endgroup$– Martin SleziakDec 23, 2016 at 15:29
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3 Answers
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Hint:
$$(1+2+3+\cdots+n)\frac{1}{n^2+n} \leq \frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}\leq (1+2+3+\cdots+n)\frac{1}{n^2+1} $$
Limit is $\frac{1}{2}$
answered Dec 15, 2015 at 15:42
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HINT:
If you are familiar with the harmonic numbers, you can also proof:
$$\lim_{n\to\infty}\frac{1}{1+n^2}+\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}=\lim_{n\to\infty}\sum_{m=1}^{n}\frac{m}{m+n^2}=$$ $$\lim_{n\to\infty}n\left(n\text{H}_{n^2}-n\text{H}_{n^2+n}+1\right)=\frac{1}{2}$$
answered Dec 15, 2015 at 17:22
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Your expression may be written as $$\lim_{n\to \infty}\sum_{k=1}^{n}\frac{k}{k+n^2}$$ so, you can search for terms $a(k,n)$ and $b(k,n)$ such that $$a(k,n)\le \frac{k}{k+n^2}\le b(k,n)$$ For example $$a(k,n)=\frac{k}{n+n^2} \text{ or } a(k,n)=\frac{k}{2n^2}$$ and $$b(k,n)=\frac{k}{k+k^2} \text{ or } b(k,n)=\frac{k}{n^2}$$
