Prove This Inequality ${\pi \over 2} \le \sum_{n=0}^{\infty} {1 \over {1+n^2}} \le {\pi \over 2} + 1$

Question

$${\pi \over 2} \le \sum_{n=0}^{\infty} {1 \over {1+n^2}} \le {\pi \over 2} + 1$$

I see I should use Riemann sum, and that $$\int_0^{\infty} {dx \over {1+x^2}} \le \sum_{n=0}^{\infty} {1 \over {1+n^2}}$$

But how do I explain this exactly, and how to get the other half of the inequality (with ${\pi \over 2} + 1$)?

Answer 1

Hint: $$ \overbrace{\sum_{n=1}^\infty\frac1{1+n^2}}^{S-1}\le\overbrace{\int_0^\infty\frac{\mathrm{d}x}{1+x^2}\vphantom{\sum_{n=0}^\infty}}^{\pi/2}\le\overbrace{\sum_{n=0}^\infty\frac1{1+n^2}}^S $$

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$\displaystyle\int_0^\infty\frac{\mathrm{d}x}{1+x^2}\le\sum_{n=0}^\infty\frac1{1+n^2}$

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$\displaystyle\sum_{n=1}^\infty\frac1{1+n^2}\le\int_0^\infty\frac{\mathrm{d}x}{1+x^2}$

Answer 2

With the same reason: $$\sum_{n=0}^{\infty} {1 \over {1+n^2}} \le \int_0^{\infty} {dx \over {1+(x-1)^2}}$$

$$\sum_{n=0}^{\infty} {1 \over {1+n^2}} \le \arctan(\infty)-\arctan(-1)=\frac{\pi}2+\frac{\pi}4 \le \frac{\pi}2+1$$

Answer 3

Hint. For each $n=0,1,2,\cdots$, you have, for all $x \in [n,n+1]$, $$ n^2+1\leq 1+x^2\leq (n+1)^2+1 $$ giving $$ \frac1{(n+1)^2+1}\leq \frac1{x^2+1}\leq \frac1{n^2+1} $$ then integrating with respect to $x$ from $n$ to $n+1$, $$ \int_n^{n+1}\frac1{(n+1)^2+1}\:dx\leq \int_n^{n+1}\frac1{x^2+1}\:dx\leq \int_n^{n+1}\frac1{n^2+1}\:dx, $$ observe that $$ \int_n^{n+1}\frac1{(n+1)^2+1}\:dx=\frac1{(n+1)^2+1},\quad \int_n^{n+1}\frac1{n^2+1}\:dx=\frac1{n^2+1} $$ giving $$ \frac1{(n+1)^2+1}\leq \int_n^{n+1}\frac1{x^2+1}\:dx\leq \frac1{n^2+1} $$ then sum from $n=0$ to $+\infty$, and with a change of indice on the left, you may conclude easily using $$ \int_0^\infty\frac1{x^2+1}\:dx=\lim_{x \to +\infty}\arctan x-\arctan 0=\frac{\pi}2. $$

Answer 4

Hint You can try by using the Taylor series of arctan evaluated at x=1

Answer 5

Hint: $$ \sum_{n=0}^{\infty} {1 \over {1+n^2}}=\frac{1}{2}+\frac{\pi}{2}\coth(\pi) $$

$$\coth(\pi)> 1\simeq 1$$ so $$\sum_{n=0}^{\infty} {1 \over {1+n^2}}\simeq\frac{1}{2}+\frac{\pi}{2}$$