Post History
#3: Post edited
by
Snoopy
·
2024-01-20T14:57:43Z (over 2 years ago)
- Without loss of generality, one can assume that $m>0$. For each positive integer $n$, since the function $x\mapsto\frac{m}{x^2+m^2}$ is decreasing, one has
- $$
\frac{m}{n^2+m^2}<\int_{n-1}^{n}\frac{m}{x^2+m^2}\ dx- $$
- It follows that
- $$
- \begin{align}
- \sum_{n=1}^\infty\frac{m}{n^2+m^2}
- &=\frac{m}{1^2+m^2}+\sum_{n=2}^\infty\frac{m}{n^2+m^2}\\\\
- &<\int_{0}^{1}\frac{m}{x^2+m^2}\ dx
- +\sum_{n=2}^\infty\int_{n-1}^{n}\frac{m}{x^2+m^2}\ dx\\\\
- &=\int_0^\infty\frac{m}{x^2+m^2}\ dx = \frac{\pi}{2}
- \end{align}
- $$
- where the last integral is found by a change of variable $u=x/m$.
- Without loss of generality, one can assume that $m>0$. For each positive integer $n$, since the function $x\mapsto\frac{m}{x^2+m^2}$ is decreasing, one has
- $$
- \frac{m}{n^2+m^2} = \int_{n-1}^{n}\frac{m}{n^2+m^2}\ dx
- <\int_{n-1}^{n}\frac{m}{x^2+m^2}\ dx
- $$
- It follows that
- $$
- \begin{align}
- \sum_{n=1}^\infty\frac{m}{n^2+m^2}
- &=\frac{m}{1^2+m^2}+\sum_{n=2}^\infty\frac{m}{n^2+m^2}\\\\
- &<\int_{0}^{1}\frac{m}{x^2+m^2}\ dx
- +\sum_{n=2}^\infty\int_{n-1}^{n}\frac{m}{x^2+m^2}\ dx\\\\
- &=\int_0^\infty\frac{m}{x^2+m^2}\ dx = \frac{\pi}{2}
- \end{align}
- $$
- where the last integral is found by a change of variable $u=x/m$.
#2: Post edited
by
Snoopy
·
2024-01-20T13:27:36Z (over 2 years ago)
- ----
- Without loss of generality, one can assume that $m>0$. For each positive integer $n$, since the function $x\mapsto\frac{m}{x^2+m^2}$ is decreasing, one has
- $$
- \frac{m}{n^2+m^2}<\int_{n-1}^{n}\frac{m}{x^2+m^2}\ dx
- $$
- It follows that
- $$
- \begin{align}
- \sum_{n=1}^\infty\frac{m}{n^2+m^2}
- &=\frac{m}{1^2+m^2}+\sum_{n=2}^\infty\frac{m}{n^2+m^2}\\\\
- &<\int_{0}^{1}\frac{m}{x^2+m^2}\ dx
- +\sum_{n=2}^\infty\int_{n-1}^{n}\frac{m}{x^2+m^2}\ dx\\\\
- &=\int_0^\infty\frac{m}{x^2+m^2}\ dx = \frac{\pi}{2}
- \end{align}
$$
- ----
- Without loss of generality, one can assume that $m>0$. For each positive integer $n$, since the function $x\mapsto\frac{m}{x^2+m^2}$ is decreasing, one has
- $$
- \frac{m}{n^2+m^2}<\int_{n-1}^{n}\frac{m}{x^2+m^2}\ dx
- $$
- It follows that
- $$
- \begin{align}
- \sum_{n=1}^\infty\frac{m}{n^2+m^2}
- &=\frac{m}{1^2+m^2}+\sum_{n=2}^\infty\frac{m}{n^2+m^2}\\\\
- &<\int_{0}^{1}\frac{m}{x^2+m^2}\ dx
- +\sum_{n=2}^\infty\int_{n-1}^{n}\frac{m}{x^2+m^2}\ dx\\\\
- &=\int_0^\infty\frac{m}{x^2+m^2}\ dx = \frac{\pi}{2}
- \end{align}
- $$
- where the last integral is found by a change of variable $u=x/m$.
#1: Initial revision
by
Snoopy
·
2024-01-20T13:26:47Z (over 2 years ago)
---
Without loss of generality, one can assume that $m>0$. For each positive integer $n$, since the function $x\mapsto\frac{m}{x^2+m^2}$ is decreasing, one has
$$
\frac{m}{n^2+m^2}<\int_{n-1}^{n}\frac{m}{x^2+m^2}\ dx
$$
It follows that
$$
\begin{align}
\sum_{n=1}^\infty\frac{m}{n^2+m^2}
&=\frac{m}{1^2+m^2}+\sum_{n=2}^\infty\frac{m}{n^2+m^2}\\\\
&<\int_{0}^{1}\frac{m}{x^2+m^2}\ dx
+\sum_{n=2}^\infty\int_{n-1}^{n}\frac{m}{x^2+m^2}\ dx\\\\
&=\int_0^\infty\frac{m}{x^2+m^2}\ dx = \frac{\pi}{2}
\end{align}
$$