Sum of $1/n+1/(n-2) + 1/(n-4) + \cdots $

Question

How does one calculate $$\frac{1}{n} + \frac{1}{n-2} + \frac{1}{n-4} \cdots $$ where this series continues until denominator is no longer positive?

$n$ is some fixed constant positive integer.

Answer 1

In the case that $n$ is even, this is a harmonic series

$\sum_{i=1}^{n/2} \frac{1}{2i}=\frac{1}{2}\sum_{i=1}^{n/2} \frac{1}{i}$.

There is no analytical solution for this sum, but the estimate

$\frac{1}{2}\ln(n/2+1)< \frac{1}{2}\sum_{i=1}^{n/2} \frac{1}{i} < \frac{1}{2}\ln(n/2)+1$

holds. If $n$ is odd, you get similar estimates using

$\int_0^{\frac{n+1}2}\frac{1}{2i-1}\ di\leq \sum_{i=1}^{(n+1)/2}\frac{1}{2i-1}\leq \int_1^{\frac{n+1}2+1}\frac{1}{2i-1}\ di$.

Answer 2

The answer will not be simple.

How to find the sum of this series : $1+\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}$

If one could find a simple expression for your sum, one could also find a simple expression for harmonic numbers and vice versa.