# Comparison test, proving divergence

If $a_n>0$ for all n, and $\sum_{n=1}^\infty a_n$ diverges, then can I prove $\sum_ {n=1}^\infty \frac{a_n}{a_n+1}$ diverges by showing that $\frac{a_n}{a_n+1}$ = $1-\frac{1}{a_n+1}$ and then use the comparison test?

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If $(a_n)$ contains infinitely many terms with $a_n > 1$, then clearly $a_n / (1 + a_n) > \frac{1}{2}$ for such $n$ and hence the second series diverges. Otherwise we may assume $a_n \leq 1$, and then we have $a_{n}/(1 + a_n) \geq \frac{1}{2} a_n$. This enables us to compare two series. – Sangchul Lee Nov 14 '12 at 12:40
Comparison is clearer with the unmodified form. – André Nicolas Nov 14 '12 at 12:45
In both cases, $a_n$ does not bound $\frac{a_n}{a_n+1}$ from above, right? So how can I use the comparison test in this case? – Alti Nov 14 '12 at 12:48

First prove that $\frac{a_n}{a_n+1}\xrightarrow[n\to\infty]{} 0\iff a_n\xrightarrow[n\to\infty]{}0$ and then use limit comparison test to show that
$\sum_{n=1}^\infty a_n$ converges iff $\sum_ {n=1}^\infty \frac{a_n}{a_n+1}$ converges:$$\displaystyle{\lim_{n\to \infty}\dfrac{\frac{a_n}{1}}{\frac{a_n}{a_n+1}}}=\lim_{n\to \infty}a_n+1=1 \neq 0.$$
Nice answer but I think your last line should actually be the first thing you say. If $a_n$ does not converge to $0$ then both diverge. WLOG assume $a_n$ converges to $0$. Then we can use the limit comparison test... – Seth Nov 14 '12 at 13:06