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Hello i'm trying to prove the following: if $\{a_n\}$s positive for all $n $ then: $\sum_{n=1}^{\infty} \frac{a_n}{a_n+1}$ iff $\sum_{n=1}^{\infty}a_n$converges.

The reverse part is pretty trivial, i'm stuck at the straight part. I know that i need to prove that $a_n \to 0$, so that the comparison test will work out. I'm trying to do that with couchy but i fail. Can anybody give me a hint? Thanks(i don't know how to write with special symbols).

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    $\begingroup$ See this $\endgroup$ Commented Dec 7, 2016 at 15:05

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Since $\displaystyle \sum_n \frac{a_n}{a_n+1}$ converges, the sequence $\displaystyle \frac{a_n}{a_n+1}$ converges to $0$, hence $a_n$ converges to $0$ (have a look at the function $x\mapsto \frac x{x+1}$).

In particular, $a_n$ is bounded by some $M\geq 0$. Hence $\frac{a_n}{a_n+1}\geq \frac{1}{1+M} a_n$.

Comparison test yields convergence of $\sum_n \frac{1}{1+M} a_n$, that is to say convergence of $\sum a_n$.

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  • $\begingroup$ Thanks, but can we prove by definition that an->0? Without this transformation? $\endgroup$ Commented Dec 7, 2016 at 13:59

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