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I have the following statement -

$$\frac{\pi}{2} < \sum_{n=0}^\infty \dfrac{1}{n^2 + 1} < \frac{3\pi}{2} $$

So I tried to prove this statement using the integral test and successfully proved the lower bound. But when I tried to calculate the upper bound I was required to calculate the integral from -1 - $\int_{-1}^{\infty} \frac{1}{x^2 +1}\,dx$. If someone can explain why it will be great , thanks!

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Note that $$ \sum_{n=0}^\infty \dfrac{1}{n^2 + 1}= \sum_{n=0}^\infty\int_{n}^{n+1}\frac{1}{n^2+1}dx>\sum_{n=0}^\infty\int_{n}^{n+1}\frac{1}{x^2+1}dx=\int_{0}^\infty\frac{1}{x^2+1}dx=\frac{\pi}{2}. $$ Also note $$ \sum_{n=0}^\infty \dfrac{1}{n^2+1}=1+\sum_{n=1}^\infty \dfrac{1}{n^2 + 1}<1+\sum_{n=1}^\infty \dfrac{1}{n^2}=1+\frac{\pi^2}{6}<\frac{3\pi}{2}. $$ Thus $$ \frac{\pi}{2}<\sum_{n=0}^\infty \dfrac{1}{n^2+1}<\frac{3\pi}{2}. $$

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  • $\begingroup$ Im sorry but i don't understand the first inequality. Could someone explain why the integral is smaller than the sum? It seems to me that if the integral is like a continous sum of points, it should be bigger than the sum of discrete numbers. Why is it different here? $\endgroup$ – Fede Poncio Jun 24 '16 at 13:55
  • $\begingroup$ @FedePoncio, In fact, $\frac{1}{n^2+1}=\int_{n}^{n+1}\frac{1}{n^2+1}dx\ge\int_{n}^{n+1}\frac{1}{x^2+1}dx$. $\endgroup$ – xpaul Jun 24 '16 at 19:41
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We have:

$$ -1+2\sum_{n\geq 0}\frac{1}{n^2+1}=\sum_{n\in\mathbb{Z}}\frac{1}{n^2+1}\color{red}{=}\sum_{n\in\mathbb{Z}}\pi e^{-2\pi|n|}=\pi\coth(\pi) \tag{1}$$

where $\color{red}{=}$ holds by the Poisson summation formula. Since $\coth(\pi)>\pi$ but $\coth(\pi)<\pi+e^{-\pi}$,

$$ \frac{\pi+1}{2}<\sum_{n\geq 0}\frac{1}{n^2+1}<\frac{\pi+1+e^{-\pi}}{2}.\tag{2}$$

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Maybe it is interesting to see that we can find directly a closed form for this series since holds $$\sum_{n\geq0}\frac{1}{1+n^{2}x^{2}}=\frac{1}{2}+\frac{\pi}{2x}\textrm{coth}\left(\frac{\pi}{x}\right) $$ hence $$\sum_{n\geq0}\frac{1}{1+n^{2}}=\frac{1}{2}\left(1+\pi\coth\left(\pi\right)\right)\approx2.0767.$$

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On the interval $(-1,0)$ is decreasing, so this won't work. Instead, we need to say that the upper bound is $\frac{1}{0^1+1}=1$ on the interval $(-1, 0)$ and then add that to the integral from $0$ to $\infty$, so we get: $$\int_{-1}^0 \frac{1}{0^1+1}dx+\int_0^\infty \frac{1}{x^2+1}dx$$ This should give you $\frac{3\pi}{2}$ as an upper bound.

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  • $\begingroup$ Ok i got it , thanks! $\endgroup$ – GeorgeB Jun 22 '16 at 13:08
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For a decreasing function $f$ on $[0,\infty)$, we have $$ f(0)+f(1)+\dots+f(m-1)\ge\int_0^m f(t)\,dt $$ Since $f(t)=\frac{1}{t^2+1}$ has $$ f'(t)=\frac{-2t}{(t^2+1)} $$ which is negative for $t>0$, we can conclude $$ \sum_{n=0}^{m-1}\frac{1}{n^2+1} \ge \int_0^{m} \frac{1}{t^2+1}\,dt $$ and, passing to the limit $$ \sum_{n=0}^{\infty}\frac{1}{n^2+1}\ge \int_{0}^\infty\frac{1}{t^2+1}\,dt=\frac{\pi}{2} $$ Similarly, we have $$ f(1)+f(2)+\dots+f(m)\le\int_0^m f(t)\,dt $$ so $$ \sum_{n=1}^{\infty}\frac{1}{n^2+1}\le\frac{\pi}{2} $$ and therefore $$ \sum_{n=0}^{\infty}\frac{1}{n^2+1}\le1+\frac{\pi}{2}<\frac{3\pi}{2} $$

You see that it was sufficient to remove the first term, in order to apply the integral upper bound.

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