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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.

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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 $\displaystyle\sum_{n=1}^{\infty}a_n \; \text{is convergent}$

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    @Peter Tamaroff: From our arguments we can show that if $a_n\geq 0\; (n\in\mathbb{N})$ then \displaystyle\sum_{n=1}^{\infty}a_n<\infty \; \Longleftrightarrow\; \displaystyle\sum_{n=1}^{\infty}\frac{a_n}{1+a_n}<\infty2012-10-16
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Assume that $ \sum\limits_{n=1}^\infty \frac{a_n}{1+a_n}<+\infty $ then $ \lim\limits_{n\to\infty}\frac{a_n}{1+a_n}=0 $ This implies $\displaystyle\lim\limits_{n\to\infty}a_n=0$ and since $a_n\geq 0$ for all $n$ we get $N\in\mathbb{N}$ such that $|a_n|<1$ for all $n>N$. For such $n$ we have $ a_n<\frac{2a_n}{1+a_n} $ As the consequence $ \sum\limits_{n=N}^\infty a_n< 2\sum\limits_{n=N}^\infty \frac{a_n}{1+a_n}<+\infty $ The rest is clear.