# Is it true that $\sum_{k=1}^n \frac{2^k}{k(k+1)} = O \left( \frac{2^n}{n(n+1)} \right)$?

I wish to know whether it is true that $$\sum_{k=1}^n \frac{2^k}{k(k+1)} = O \left( \frac{2^n}{n(n+1)} \right) , \quad n \to \infty.$$ That is, whether the left-hand side is bounded by some constant $C$ times $\frac{2^n}{n(n+1)}$ for large enough $n$. Intuitively, I am inclined to say that the growth of the left-hand side should be "dictated" by the final term in the summation, since successive terms are growing a bit slower than exponentially. However, after several attempts in vain to bound the growth of this summation by the final term, I'm wondering if the statement above is actually true. Any hints or ideas? Thanks in advance!

• "after several attempts in vain to bound the growth of this summation by the final term" Please show one or two of these attempts.
– Did
Commented Nov 1, 2017 at 23:15

$$\begin{eqnarray*}\sum_{k=1}^{n}\frac{2^k}{k(k+1)}&\stackrel{\text{SBP}}{=}&\frac{2(2^n-1)}{n(n+1)}-\sum_{k=1}^{n-1}2(2^k-1)\left[\frac{1}{(k+1)(k+2)}-\frac{1}{k(k+1)}\right]\\&\leq&2\cdot\frac{2^n}{n(n+1)}+4\sum_{k=1}^{n-1}\frac{2^k}{k(k+1)(k+2)}\\&\leq&2\cdot\frac{2^n}{n(n+1)}+\frac{38}{15}+\frac{4}{6}\sum_{k=4}^{n-1}\frac{2^k}{k(k+1)}\end{eqnarray*}$$ leads to $\sum_{k=1}^{n}\frac{2^k}{k(k+1)}\leq 7\cdot\frac{2^n}{n(n+1)}$ for any $n\geq 5$. $\text{SBP}$ stands for summation by parts.