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Given that $μ$ and $Q$ are real constants and $i$ is a positive integer, evaluate

$\sum_{i=1}^{+\infty}\;i\,\tan^{-1}\left(\frac{\mu Q}{\mu^2+\left(i^2-\frac{1}{4}\right)Q^2}\right)$

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    Hmm. Try and find where it might converge. It seems to only do so when $\mu$ or $Q$ is zero.2011-10-23

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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?

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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.

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    Thanks, for continuous sequence ∫xp(x)dx from x=0..∞ is ∞, so.. for given Q and μ, so it may be safe to assume its quantized version is also unbounded. In fact for larger values of x, xp(x) closely resembles harmonic series!2011-10-24