# Prove $\sum_{k=0}^{\infty} \frac {\log(1+k)}{2^k} \le \log3$

Initially I faced with a question $$\sqrt{2\sqrt{3\sqrt{4...\sqrt{N}}}} < 3$$ which is not hard to prove by induction.

But if taking logarithm on both sides of this inequality, we get $$\sum_{k=0}^{\infty} \frac{\log(1+k)}{2^k} < \log3$$.

Is there any way to prove this series inequality? Or evaluate this series?

I tried to use series of $$\log(1+k)$$. And I got $$\sum_{k=0}^{\infty} \frac{\log(1+k)}{2^k} < \sum_{k=0}^{\infty} \frac{k}{2^k} =2$$, which gives a larger upper bound.

A surprising fact is that the more term we use for the series of $$\log(1+k)$$, the larger upper bound we will get.

• The "surprising" fact is actually reasonable, since the series $\log(1+k)$ does not converge: $\log(1+x)$ could be expanded at $x=0$ in the interval $(-1,1)$, out of this interval the series would diverge.
– xbh
Oct 26, 2018 at 3:44

It is easy to show that $$n^3 \leq 3^n$$ for all $$n \in \Bbb{N}$$. Thus $$n \leq 3^{\frac{n}{3}}$$, i.e., $$\log n \leq \frac{n}{3} \log 3$$. Hence, $$\sum\limits_{k=1}^\infty \frac{\log(1+k)}{2^k} \leq \sum\limits_{k=1}^\infty \left(\frac{1+k}{3 \cdot 2^k}\right) \log 3 =\log 3$$

• Very nice solution! Is there any motivation behind this? $\sum_{k=1}^{\infty}\frac{1+k}{3 \cdot 2^k}=1$ just perfectly works here. Oct 26, 2018 at 6:13
• @NagiYang, thank you! If $\frac{\log(1+n)}{2^n} \leq \frac{\log 3}{2^n}$, then the inequality is strongly true. But obviously this inequality is false. Hence, I think of a function $f$ satisfying $\frac{\log(1+n)}{2^n} \leq f(n) \frac{\log 3}{2^n}$, and I found that $f(n)=\frac{1+n}{3}$ is well operated. Then I want the inequality $\sum\frac{1+k}{3\cdot 2^k} \leq 1$, and when calculating, the equality holds. Oct 26, 2018 at 10:30

Hint: Once you have an upper bound, you can try using a couple terms of the actual sequence and then using the bound. So, for any positive integer $$N$$,

$$\sum_{k=0}^{\infty} \frac{\log(1+k)}{2^k} =\sum_{k=0}^N \frac{\log(1+k)}{2^k} + \sum_{k=N+1}^{\infty} \frac{\log(1+k)}{2^k} < \sum_{k=0}^N\frac{\log(1+k)}{2^k} + \sum_{k=N+1}^{\infty} \frac{k}{2^k}$$

What about summation by parts? If we let $$S=\sum_{k\geq 0}\frac{\log(k+1)}{2^k}$$ we have

$$\begin{eqnarray*}\sum_{k=0}^{N}\frac{\log(k+1)}{2^k} &=& (2-2^{-N})\log(N+1)-\sum_{k=0}^{N-1}(2-2^{-k})\log\left(\frac{k+2}{k+1}\right)\\ &=&-2^{-N}\log(N+1)+\sum_{k=0}^{N-1}2^{-k}\log\left(1+\frac{1}{k+1}\right)\end{eqnarray*}$$ hence by letting $$N\to +\infty$$ we get $$\color{red}{S} = 2\sum_{k\geq 1}\frac{\log\left(1+\frac{1}{k}\right)}{2^k}\color{red}{<}\log(2)+2\sum_{k\geq 2}\frac{1}{k 2^k}={3\log(2)-1}\approx \color{red}{1.07944}$$ which is a better bound, equivalent to $$\sqrt{2\sqrt{3\sqrt{4\sqrt{\ldots}}}}<\frac{8}{e}.$$

Using $$\log(1+x)> \frac{x}{1+\frac{x}{2}}$$ on $$(0,1)$$, we also get the tight lower bound $$\color{red}{S >} -\frac{14}{3}+\log(2)+4\sqrt{2}\log(1+\sqrt{2})\approx \color{red}{1.01228}.$$