3
$\begingroup$

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

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

  • 1
    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.2011-10-23
  • 0
    @anon: Sorry! "i" runs over an infinite number of values and NOT imaginary unit2011-10-23
  • 0
    Hmm. Try and find where it might converge. It seems to only do so when $\mu$ or $Q$ is zero.2011-10-23

2 Answers 2

2

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?

1

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.

  • 0
    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)2011-10-24
  • 0
    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