Given that $μ$ and $Q$ are real constants and $i$ is a positive integer, evaluate
$\sum_{i=1}^{+\infty}\;i\,\tan^{-1}\left(\frac{\mu Q}{\mu^2+\left(i^2-\frac{1}{4}\right)Q^2}\right)$
Given that $μ$ and $Q$ are real constants and $i$ is a positive integer, evaluate
$\sum_{i=1}^{+\infty}\;i\,\tan^{-1}\left(\frac{\mu Q}{\mu^2+\left(i^2-\frac{1}{4}\right)Q^2}\right)$
Rewrite the expression as follows:
$\displaystyle \sum_{i=1}^\infty i \left[\tan^{-1} \frac{Q}{\mu}(i+\frac{1}{2}) - \tan^{-1} \frac{Q}{\mu}(i-\frac{1}{2})\right]$
Can you find the sum now?
Arctan is bounded; as $i$ increases, the fraction gets smaller, so overall $\arctan(\ldots)$ approaches 0. The question is, does it go to zero faster than $i$ goes to $\infty$? If so, you still need to consider the summation, and determine whether it converges or not. Try comparing it with a series that doesn't converge, like the harmonic series
$ \underset{n=1}{\overset{\infty}{\sum}}\ \frac{1}{n} $
If you can bound your sequence from below by a divergent series, then your series diverges by the comparison test. It converges if you can bound it above by a convergent series.