Problem 169
Easy Numerical and Functional Series

Determine the region of absolute and conditional

Determine the region of absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty} \frac1{x^2+n^2}. $$

Explanation of Method

Since $x^2\ge0$, every term satisfies $0<\frac1{x^2+n^2}\le\frac1{n^2}$, and the convergent $p$-series $\sum1/n^2$ immediately gives convergence by direct comparison, for every real $x$ simultaneously. Because all terms are already positive, absolute convergence and ordinary convergence coincide.

  1. Bound each term above by $1/n^2$.
  2. Apply the comparison test against the convergent $p$-series $\sum1/n^2$.
  3. Note that positivity of all terms makes convergence automatically absolute.

Step 1 · Bound the terms

For any real $x$ and $n\ge1$: $\dfrac1{x^2+n^2}\le\dfrac1{n^2}$, since $x^2\ge0$.

Step 2 · Comparison test

$\sum_{n=1}^{\infty}1/n^2$ converges ($p$-series, $p=2>1$), so by direct comparison $\sum_{n=1}^{\infty}1/(x^2+n^2)$ converges for every $x\in\mathbb R$.

Step 3 · Absolute convergence is automatic

Since every term is positive, $\left|\dfrac1{x^2+n^2}\right|=\dfrac1{x^2+n^2}$, so convergence and absolute convergence coincide — the series converges absolutely for all real $x$. At $x=0$: $\sum1/n^2=\pi^2/6$ (the Basel problem); for general $x\ne0$, the closed form is $\dfrac{\pi\coth(\pi|x|)-1/|x|}{2|x|}$ (via the classical partial-fraction expansion of $\coth$).


Summary Table

RegionBehavior
All $x\in\mathbb R$Absolutely convergent
Special case $x=0$Sum $=\pi^2/6$

Additional Notes

Result

The series converges absolutely for every real $x$:

$$ \sum_{n=1}^{\infty} \frac1{x^2+n^2} \text{ converges absolutely for all } x\in\mathbb R. $$