For $a_n$ positive sequence.
I think I can prove one direction, but not both.
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Which direction? – Buzi May 17 '15 at 13:00
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Any assumptions on $a_n$? (are they complex numbers, real numbers, positive numbers?) – Wojowu May 17 '15 at 13:01
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Which direction you proved? How? – wythagoras May 17 '15 at 13:01
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@Wojowu real numbers – Yoav R. May 17 '15 at 13:01
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1@YoavR. In this case $a_n=-2$ would form a counterexample. I believe you intend the numbers to be positive as well. – Wojowu May 17 '15 at 13:03
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$ \sum_{n=1}^\infty a_n<\infty \implies \sum_{n=1}^\infty \frac{a_n}{1+a_n} <\infty$ - this is pretty straight forward as $a_n > \frac{a_n}{1+a_n}$ – Yoav R. May 17 '15 at 13:04
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1$\Rightarrow$ seems pretty obvious, just use the comparison test. So I assume it's $\Leftarrow$ you're having trouble with? – Gregory Grant May 17 '15 at 13:05
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@Wojowu - as you can see - $a_n$ positive sequence - as I already said – Yoav R. May 17 '15 at 13:06
1 Answers
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Since $\frac{a_n}{1+a_n}<a_n$, thus $\sum_{n=1}^\infty a_n<\infty$ implies $\sum_{n=1}^\infty \frac{a_n}{1+a_n}<\infty$. On the other hand, let $\sum_{n=1}^\infty \frac{a_n}{1+a_n}<\infty$. Then $\lim_{n\to\infty}\frac{a_n}{1+a_n}=0$ and hence $\lim_{n\to\infty}a_n=0$. So there is $N>0$ such that $a_n<1$ when $n\ge N$. From this, we have $$ \frac{1}{2}a_n\le\frac{a_n}{1+a_n}, \text{ for }n\ge N $$ which implies $\sum_{n=1}^\infty a_n<\infty$.
xpaul
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Can't figure out this thing you wrote: "Then $\lim_{n\to\infty}\frac{a_n}{1+a_n}=0$ and hence $\lim_{n\to\infty}a_n=0$" – Yoav R. May 17 '15 at 13:12
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@YoavR., you can prove this fact that $\lim_{n\to\infty}\frac{a_n}{1+a_n}=0$ implies $\lim_{n\to\infty}a_n=0$. – xpaul May 17 '15 at 13:20
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@YoavR., ok. Clearly if $\lim_{n\to\infty}\frac{a_n}{1+a_n}=0$, then ${a_n}$ is bounded, say, $a_n\le M$ for all $n\ge 1$. Thus $\frac{1}{1+M}a_n\le\frac{a_n}{1+a_n}$. Thus $\lim_{n\to\infty}\frac{a_n}{1+a_n}=0$ implies $\lim_{n\to\infty}a_n=0$ – xpaul May 17 '15 at 13:40