# Prove $\sum _{k=1}^{\infty}\frac{a_k}{a_k+1}\:$ diverges

Let $\{ a_n \}_{n=1}^{\infty}>0$ be a sequence and $S_n=\sum _{k=1}^{n }a_k\:\rightarrow\infty$ when $n\rightarrow\infty$. Prove $\sum _{k=1}^{\infty}\frac{a_k}{a_k+1}\:$ diverges.

I tried to some algebraic manipulations with the expression need to be proved, but got stuck.

Any help appreciated.

• Hint: $$\frac{a_k}{a_k+1}=1-\frac1{a_k+1}$$ – vrugtehagel Mar 7 '17 at 8:35
• If $a_n \gt 0$ then $\dfrac{a_n}{a_n+1} \ge \min\left(\dfrac12, \dfrac12 a_n\right)$ – Henry Mar 7 '17 at 8:53

It suffices to consider $a_k$ going to zero. Since $a_k$ is non-increasing and positive, we could apply Cauchy condensation test which states: the series \begin{align} \sum^\infty_{k=1} a_k \end{align} diverges if and only if \begin{align} \sum^\infty_{k=1}2^k a_{2^k} \end{align} diverges.
Now, observe for \begin{align} \sum^\infty_{k=1}2^k\frac{a_{2^k}}{1+a_{2^k}}\geq \sum^N_{k=1}2^k\frac{a_{2^k}}{1+a_{2^k}} + \frac{1}{2}\sum^\infty_{k=N+1}2^k a_{2^{k}} = \infty. \end{align} Note, since $a_k \rightarrow 0$, then it follows there exists $N$ such that $a_{2^k}<1$.
If $a_n$ is bounded with $a_n < A$, then $a_n/(1 +a_n) > a_n/(1 + A)$ and the series diverges by comparison with $\sum a_n$.
If $a_n$ is unbounded then there is a subsequence $a_{n_k} \to \infty$. Hence $a_{n_k}/(1 + a_{n_k}) \to 1$ and $\lim_{n \to \infty} a_n/(1 + a_n) \neq 0.$