TITLE: If $\{a_n\}$ is not summable, neither is $\left\{ {\frac{{{a_n}}}{{1 + {a_n}}}} \right\}$
QUESTION [5 upvotes]: Let $\{a_n\}$ be a sequence. We say $\{a_n\}$ is summable if the sequence $\{s_n\}$ defined by $$s_1=a_1\\s_{n+1}=s_n+a_{n+1}$$
converges to $\ell\in\Bbb R$, and write $$\sum\limits_{n = 1}^\infty  {{a_n}}  = \ell $$

$(1)$ Suppose  $\{a_n\}$ is of non-negative terms, and that it is not summable, that is $\lim {s_n}$ fails to exist. Show that $$\left\{ {\frac{{{a_n}}}{{1 + {a_n}}}} \right\}$$ is not summable either.

Now, since $$0 \leqslant \frac{{{a_n}}}{{1 + {a_n}}} \leqslant {a_n}$$ for each $n$, $\{a_n\}$ being summable implies $\left\{ {\dfrac{{{a_n}}}{{1 + {a_n}}}} \right\}$ also is (monotone convergence). I need to show the converse of this, but I can see no way to prove it. Since $a_n\geq0$, the partial sums $${s_N} = \sum\limits_{n = 1}^N {{a_n}} $$can be made as large as we want.

REPLY [14 votes]: Suppose that $a_n\geq 0 (n\in \mathbb{N})$ and
$\displaystyle\sum_{n=1}^{\infty}\frac{a_n}{1+a_n} \; \text{is convergent}$.
Then 
$$
\lim_{n\rightarrow\infty}\frac{a_n}{1+a_n}=0.
$$
It implies that $\displaystyle\lim_{n\rightarrow\infty}a_n=0$ and
$$
\lim_{n\rightarrow\infty}\frac{a_n}{\frac{a_n}{1+a_n}}=\lim_{n\rightarrow\infty}(1+a_n)=1.
$$
Since $\displaystyle\sum_{n=1}^{\infty}\frac{a_n}{1+a_n} \; \text{is convergent}$, we have 
$\displaystyle\sum_{n=1}^{\infty}a_n \; \text{is convergent}$