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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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  • $\begingroup$ Nice one. ${}{}$ $\endgroup$ – Pedro Tamaroff Oct 16 '12 at 1:52
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    $\begingroup$ @Peter Tamaroff: From our arguments we can show that if $a_n\geq 0\; (n\in\mathbb{N})$ then $$\displaystyle\sum_{n=1}^{\infty}a_n<\infty \; \Longleftrightarrow\; \displaystyle\sum_{n=1}^{\infty}\frac{a_n}{1+a_n}<\infty$$ $\endgroup$ – blindman Oct 16 '12 at 2:13
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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.

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