Evaluate $\sum_{k=1}^n\log(x ^ {\frac{k}{2k+1}} )$

Question

I have absolutely no idea on how to tackle this. The log is completely throwing me off, so if you could explain how to deal with a log in a series i would really appreciate it!

Answer 1

Following comments, you need to evaluate $$S_n=\sum_{k=1}^n\frac{k}{2k+1}$$

What is easy is to show that $$S_n=\sum_{k=1}^n\frac{k}{2k+1}\lt \sum_{k=1}^n\frac{k}{2k+0}=\sum_{k=1}^n\frac{1}{2}=\frac{n}{2}$$ which is then an upper bound.

Going beyond is much more complex since $$S_n=\frac{n}{2}+\frac{1}{4} \left(\psi ^{(0)}\left(\frac{3}{2}\right)-\psi ^{(0)}\left(n+\frac{3}{2}\right)\right)$$ where appears the digamma function.

If $n$ is large, you can use the asymptotics to get $$S_n =\frac{n}{2}-\frac{1}{4}\log(n)+\frac{1}{4} \psi ^{(0)}\left(\frac{3}{2}\right)-\frac{1}{4 n}+O\left(\frac{1}{n^2}\right)$$ where $\psi^{(0)}\left(\frac{3}{2}\right)=2-\gamma -2\log (2)\approx 0.03649$.

Using $n=10$, the exact result would be $\frac{64157087}{14549535}\approx 4.40956$ while the above expansion would give $\frac{219}{40}-\frac{1}{4} (\gamma +\log (40))\approx 4.40848$.

Answer 2

I am pretty sure there is a typo in your expression, but notice how $log(x^a)=alog(x)$ and $\sum \frac{k}{2k+1} log(x) = log(x) \sum \frac{k}{2k+1} = log(x) ( \sum \frac{1}{2} (1 - \frac{1}{2k+1})) $.

Answer 3

Notice this can be written as $$\log(x)\sum\frac{k}{2k+1}$$ Since $x$ is a constant with respect to the sum and $\log(x^a)=a\log(x)$
Can you continue?