Supremum of this set of sums $\sum_{m=1}^{n}{\frac{1}{m*2^m}}$

Question

$S_n= \sum_{m=1}^{n}{\frac{1}{m*2^m}}$ if $n\ge1$

does it have a supremum?

Answer 1

You mean $S_n=\sum_{k=1}^n\frac{2^k}{k}$? Clearly, $S_n\ge \frac{2^n}{n}$. We know that $\lim_{n\to\infty}\frac{2^n}{n}=\infty$. So, no, $S_n$ has no upper bound, i.e., no supremum.

But if you mean $S_n=\sum_{k=1}^n\frac{1}{k2^k}$, then $S_n\le \sum_{k=1}^n\frac{1}{2^k}<1$. So, it has an upper bound, so $\{S_n\}$ has supremum. (In fact the supremum is $\log2$, since $\log(1-x)=-\sum_{k=1}^\infty\frac{x^k}{k}$, and we can plug in $x=\frac12$ there.)