I'm asked to prove that $$\lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)=\frac{\pi}{4}$$ This looks like it can be solved with Riemann sums, so I proceed:

\begin{align*} \lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)&=\lim_{n \to \infty} \sum_{k=1}^{n}\frac{n}{n^2+k^2}\\ &=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{n^2}{n^2+k^2})\\ &=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{1}{1+(k/n)^2})\\ &=\lim_{n \to \infty} \sum_{k=1}^{n}f(\frac{k}{n})(\frac{k-(k-1)}{n})\\ &=\int_{0}^{1}\frac{1}{1+x^2}dx=\frac{\pi}{4} \end{align*}

where $f(x)=\frac{1}{1+x^2}$. Is this correct, are there any steps where I am not clear?

  • 1
    $\begingroup$ Though the derivation is correct, you shouldn't use the $\lim$ symbol until you are sure that the limit exists. $\endgroup$
    – yoann
    Aug 16, 2015 at 14:59
  • $\begingroup$ What steps must I go through in order to show that the limit exists? $\endgroup$
    – lamyvista
    Aug 16, 2015 at 18:49
  • $\begingroup$ Usually you would perform your computations, and show that the last term has a limit. An easy way to avoid doing mistakes is to write $A_n = B_n = \dots = Z_n \longrightarrow_{n \to +\infty} \ell$. $\endgroup$
    – yoann
    Aug 16, 2015 at 19:52

2 Answers 2


Notice, we have $$\lim_{n\to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\dots +\frac{n}{n^2+n^2}\right)=\lim_{n\to \infty}\sum_{r=1}^{n}\frac{n}{n^2+r^2}$$

$$\lim_{n\to \infty}\sum_{r=1}^{n}\frac{n}{n^2+r^2}=\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\frac{1}{n}}{1+\left(\frac{r}{n}\right)^2}$$ Let, $\frac{r}{n}=x\implies \lim_{n\to \infty}\frac{1}{n}=dx\to 0$

$$\text{upper limit of x}=\lim_{n\to \infty }\frac{n}{n}=1$$ $$\text{lower limit of x}=\lim_{n\to \infty }\frac{1}{n}=0$$ Hence, using integration with proper limits, we get

$$\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\frac{1}{n}}{1+\left(\frac{r}{n}\right)^2}= \int_ {0}^{1}\frac{dx}{1+x^2}$$ $$=\left[\tan^{-1}(x)\right]_{0}^{1}$$ $$=\left[\tan^{-1}(1)-\tan^{-1}(0)\right]$$ $$=\left[\frac{\pi}{4}-0\right]=\frac{\pi}{4}$$

  • $\begingroup$ Is the extra dx in your integrand a typo? Thanks! $\endgroup$
    – lamyvista
    Aug 16, 2015 at 8:27
  • 3
    $\begingroup$ Thank you for observation! $\endgroup$ Aug 16, 2015 at 8:29
  • $\begingroup$ Forgive my ignorance, but could you explain how $lim_{n\to \infty}\frac{1}{n}=dx$ $\endgroup$
    – meiji163
    Aug 16, 2015 at 15:44
  • $\begingroup$ @Sky $\lim_{n\to\infty}\frac1n\ne\text{d}x$, that would be gibberish. What's happening is Harish is using the fact that $\lim_{n\to\infty}\frac1n\sum_{r=1}^nf\left(\frac{r}{n}\right)=\int_0^1f(x)\text{d}x$, which is the definition of Riemann integration. $\endgroup$ Aug 16, 2015 at 15:58

Yes, what you've written is correct.

I think your write-up would be cleaner if you explicitly mentioned the antiderivative $$ \int \frac{1}{1 + x^2} dx = \arctan x,$$ which makes it completely clear where $\frac{\pi}{4}$ comes from at the end.


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