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Sum of $$\lim_{n\rightarrow \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots \cdots \cdots +\frac{n}{n^2+n}\right)$$

$\bf{My\; Try::}$ I have solved it using Squeeze theorem

$$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+n}\leq \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+r}\leq \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+1}$$

So we get $$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+r} = 1$$

My question is can we solve above limit without using Squeeze Theorem,

If yes then plz explian me, Thanks

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    $\begingroup$ You mean, finding more complicated and less enlightening ways? What for? $\endgroup$ – Did Nov 2 '16 at 10:28
  • $\begingroup$ Use Riemann sum instead $\endgroup$ – Alex Nov 2 '16 at 10:33
  • $\begingroup$ You have the sum of $n$ terms bounded between $\frac{1}{n+1}$ and $\frac{1}{n}$, so it is trivial that the limit is $1$. What is the reason for looking for more involved approaches? $\endgroup$ – Jack D'Aurizio Nov 2 '16 at 16:53
  • $\begingroup$ This question is about the same sum: Convergence of a sequence, $a_n=\sum_1^nn/(n^2+k)$. Found using Approach0. $\endgroup$ – Martin Sleziak Dec 23 '16 at 15:26
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Using harmonic numbers $$S_n=\sum^{n}_{r=1}\frac{n}{n^2+r}=n \left(H_{n^2+n}-H_{n^2}\right)$$ Now, using the expansion $$H_p=\gamma+\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p ^4}\right)$$ and applying to each term you should arrive to $$S_n=n\left(\frac{-6 n^3-6 n^2+2 n+1}{12 n^4 (n+1)^2}+\log \left(1+\frac{1}{n}\right)+O\left(\frac{1}{n ^2}\right)\right)$$ Taylor again, $$S_n=1-\frac{1}{2 n}+O\left(\frac{1}{n ^2}\right)$$

Edit

May be of interest $$T_n=\sum^{n}_{r=1}\frac{n}{n^k+r}=n \left(H_{n^k+n}-H_{n^k}\right)$$ leads to $$T_n=\frac{ \left(n^{2-k}-6 n^{k+1}-6 n^2+2 n\right)}{12 n^k\left(n^k+n\right)^2}+\log \left(1+\frac 1 {n^{k-1}}\right)$$ which makes $$n^{k-1}T_n=1-\frac{1}{2 n^{k-1}}+O\left(\frac{1}{n^k}\right)$$

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