$\sum_{k=1}^\infty a_k$ converges iff $\sum_{k=1}^\infty a_k/(1+a_k)$ converges

I just came across the following exercise in Schinazi, R. (2012). From Calculus to Analysis (1st ed., p. 54). Basel: Birkhäuser.

Let $a_n>0$. Show that $\sum_{k=1}^\infty a_k$ converges if and only if $$\sum_{k=1}^\infty \frac{a_k}{1+a_k}$$ converges.

Just from the looks of it I assume I'd need to use the comparison test for the $\Rightarrow$ direction. What about the other direction though?

• How about limit-comparison? – Lord Shark the Unknown Jan 9 '18 at 18:42
• @LordSharktheUnknown With regards to what direction? I assume the $\Leftarrow$ direction, isn't it? – Christian Ivicevic Jan 9 '18 at 18:55
• both directions? – Lord Shark the Unknown Jan 9 '18 at 18:57
• – Martin R Jan 9 '18 at 21:35

If $\sum\frac{a_k}{1+a_k}$ converges, then necessarily $$\frac{a_k}{1+a_k}\to0\text{ as }k\to\infty,$$ in turn implying that $a_k\to0$ as $k\to\infty$. Then $a_k\leq 1$ for $k$ sufficiently large, and therefore $$\frac{a_k}{1+a_k}\geq\frac{a_k}{2}\text{ for k sufficiently large}.$$ Then, use the fact that $\sum a_k$ converges if and only if $\sum\frac{a_k}{2}$ converges.

For the reverse implication, show that if $\sum a_n$ diverges then so does $\sum a_n/(1+a_n)$.

If $a_n$ is bounded and $a_n < B$ for all $n$ then $a_n/(1+a_n) > a_n/(1+B)$ and we have divergence by the comparison test.

If $a_n >0$ is unbounded then there is a subsequence $a_{n_k} \to \infty$ and $a_{n_k}/(1 + a_{n_k}) \to 1$ implying divergence by the term test.

We will use the limit comparison test. $$\lim_{n\to \infty}\frac{a_n (1-a_n)}{a_n} = 1 > 0$$ The results follows

$\displaystyle \sum \dfrac{a_k}{1+a_k}$ converges

$\rightarrow$

$\lim_{k \rightarrow \infty} \dfrac{a_k}{1+a_k}=0,$

$\rightarrow$

$a_k$ is bounded , see note.

There is a $n_0$ such that $a_k \lt 1.$

$a_k= \dfrac{a_k(1+a_k)}{1+a_k} \lt \dfrac{a_k(1+1)}{1+a_k} =$

$\dfrac{2a_k}{1+a_k}.$

Comparison test.

Note:

$\lim_{k \rightarrow \infty} \dfrac{a_k}{1+a_k} = 0.$

For $1 > \epsilon >0$ there is a $n_0$ such that

for $k\ge n_0$ :

$\dfrac{a_k}{1+a_k} \lt \epsilon$, or

$a_k(1-\epsilon) \lt \epsilon ,$

$a_k \lt \dfrac{\epsilon}{1-\epsilon}.$

Choose $\epsilon =1/2$ to get $a_k\lt 1.$