# Any hint on this summation problem?

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

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

• 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
Commented Oct 23, 2011 at 17:37
• @anon: Sorry! "i" runs over an infinite number of values and NOT imaginary unit Commented Oct 23, 2011 at 17:47
• Hmm. Try and find where it might converge. It seems to only do so when $\mu$ or $Q$ is zero.
– anon
Commented Oct 23, 2011 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?

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.

• 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) Commented Oct 24, 2011 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! Commented Oct 24, 2011 at 1:53