How to compute the limit of $\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+\cdots+\frac{n}{n^2+n^2}$ without using Riemann sums?

My Try: I have Solved It using Limit as a Sum (Reinman Sum of Integral.)

But I did not understand How can I solve it Using Sequeeze Theorem or any other way.

  • $\begingroup$ No Jack D'Aurizio It was spelling Mistake. $\endgroup$
    – juantheron
    Aug 2, 2016 at 14:28
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    $\begingroup$ @JackD'Aurizio lol $\endgroup$
    – Aweygan
    Aug 2, 2016 at 14:29
  • $\begingroup$ Is there a particular reason behind wanting to solve this without Riemann sums or is it just a challenge you thought of? $\endgroup$
    – b00n heT
    Aug 2, 2016 at 14:32
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    $\begingroup$ The big other option, if Riemann sums are forbidden, is to compare directly these sums to integrals. For example, the function $x\mapsto n/(n^2+x^2)$ is decreasing on $x\geqslant0$ hence $$\frac{n}{n^2+k^2}\leqslant\int_{k-1}^k\frac{n}{n^2+x^2}dx$$ hence the $n$th sum is less than $$\int_0^n\frac{n}{n^2+x^2}dx$$ which, by the change of variable $x=nu$, is $$\int_0^1\frac{1}{1+u^2}du=\frac{\pi}4.$$ Likewise for the lower bound. Yes, in a sense this is cheating... but formally, one is using the squeeze principle. $\endgroup$
    – Did
    Aug 2, 2016 at 14:33
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    $\begingroup$ All in all, and unless I am mistaken, this proves that, for every $n$, $$\frac\pi4+\frac1{2n}-\arctan\frac1n\leqslant\sum_{k=1}^n\frac{n}{n^2+k^2}{}{}{}{}\leqslant\frac\pi4$$ $\endgroup$
    – Did
    Aug 2, 2016 at 14:40

2 Answers 2


For $0 \le k \le n$, let $t_k = \frac{k}{n}$.

Divide the interval $[0,1]$ into $n$ sub-intervals $[t_{k-1},t_k]$ for $1 \le k \le n$ and apply MVT to $\tan^{-1} x$ on the intervals. We find for each $k$, there is a $x_k \in ( t_{k-1}, t_k )$ such that $$\tan^{-1}t_k - \tan^{-1} t_{k-1} = \frac{t_k-t_{k-1}}{1+x_k^2}$$ Since $\frac{1}{1+x^2}$ is monotonic decreasing for $x \ge 0$, we have the bound

$$\frac{n}{n^2 + (k-1)^2} = \frac{t_k-t_{k-1}}{1+t_{k-1}^2} \ge \tan^{-1}t_k - \tan^{-1}t_{k-1} \ge \frac{t_k-t_{k-1}}{1+t_k^2} = \frac{n}{n^2+k^2}$$

Summing this over $k$ from $1$ to $n$, we get

$$\frac{1}{2n} + \sum_{k=1}^n \frac{n}{n^2+k^2} = \sum_{k=1}^n \frac{n}{n^2+(k-1)^2} \ge \tan^{-1}t_n - \tan^{-1}t_0 = \tan^{-1} 1 = \frac{\pi}{4} \ge \sum_{k=1}^n\frac{n}{n^2+k^2}$$ This leads to $\displaystyle\;\left|\sum_{k=1}^n \frac{n}{n^2+k^2} - \frac{\pi}{4}\right| \le \frac{1}{2n}$ and hence $\displaystyle\;\lim_{n\to\infty} \sum_{k=1}^n \frac{n}{n^2+k^2} = \frac{\pi}{4}$


One may use the digamma function, from the standard $ \psi(z+1)-\psi(z)=\dfrac1z$ one obtains easily $$ \psi(z+n+1)-\psi(z+1)=\sum_{k=1}^n\frac1{z+k}, $$ inserting $z:=in$ and considering imaginary parts gives $$ \sum_{k=1}^n\frac{n}{n^2+k^2}=-\text{Im}\left[\psi(in+n+1)-\psi(in+1)\right] $$ then one may recall that (see $6.3.18$ here) $$ \psi(z)=\log z+O\left(\frac1z \right), \quad z \to \infty, \quad |\mathrm{arg}z|<\pi, $$ which yields, as $n \to \infty$,

$$ \sum_{k=1}^n\frac{n}{n^2+k^2}=-\text{Im}\left[\log(1+i)-\log i \right]+O\left(\frac1n \right)=\frac\pi4+O\left(\frac1n \right) \to \frac\pi4. $$

Remark. By considering more asymptotic terms in $\psi$, one gets,

$$ \sum_{k=1}^n\frac{n}{n^2+k^2}=\frac\pi4-\frac1{4n}-\frac1{24n^2}+\frac1{2016n^6}+O\left(\frac1{n^8} \right) $$

$\displaystyle \log z$ denotes the principal value of the logarithm defined by $$ \begin{align} \displaystyle \log z = \ln |z| + i \: \mathrm{arg}z, \quad -\pi <\mathrm{arg} z \leq \pi,\quad z \neq 0. \end{align} $$

  • $\begingroup$ How do you derive that first Summation equivalence? I feel like I'm missing something. The rest is trivial. $\endgroup$ Aug 2, 2016 at 17:10
  • $\begingroup$ From the $\psi(z+1)-\psi(z)=\frac1{z}$ one deduces $\psi(z+n+1)-\psi(z+1)=\sum_{k=1}^n\frac1{z+k}$, then one may insert $z:=in$ and take minus the imaginary part. $\endgroup$ Aug 2, 2016 at 17:25

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