Home » Uncategorized » On the convergence of a series

On the convergence of a series

Let \alpha>0 and let \mathcal{H}_n denote the n-th harmonic number. Examine the convergence of the series

    \[\mathcal{S} = \sum_{n=1}^{\infty} \alpha^{\mathcal{H}_n}\]

Solution

We begin by the very well known

    \[\mathcal{H}_n = \log n + \mathcal{O}(1)\]

So \alpha^{\mathcal{H}_n} is comparable to a constant multiple of \alpha^{\log n} = n^{\log a}. Thus, the series converges if \log \alpha<-1 that is to say 0<\alpha<\frac{1}{e} and diverges otherwise.

 

Read more

Leave a comment

Donate to Tolaso Network