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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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Minor points: it's potentially confusing to use $i$ as a variable when it could also be the imaginary unit, and it's not simply "a" positive integer, because it runs over an infinite number of values. – anon Oct 23 '11 at 17:37
@anon: Sorry! "i" runs over an infinite number of values and NOT imaginary unit – Bhaskar Dey Oct 23 '11 at 17:47
Hmm. Try and find where it might converge. It seems to only do so when $\mu$ or $Q$ is zero. – anon Oct 23 '11 at 17:51

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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It is a quantized Cauchy sequence where Q represents quantization parameter and μ is the distribution parameter of Cauchy pdf p(x)=(1/π)(μ/μ2+x2) – Bhaskar Dey Oct 24 '11 at 1:28
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! – Bhaskar Dey Oct 24 '11 at 1:53

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