Let $\{a_n\}$ be a positive sequence. Prove $\sum a_n$ converges $\iff$ $\sum{a_n\over a_n+1}$ converges.

I showed $\sum a_n$ converges $\Rightarrow$ $\sum{a_n\over a_n+1}$ converges.

Proving the other side, I was looking at $\lim\inf {a_n \over({a_n\over a_n+1})}=\lim\inf (a_n+1)$ and $\lim \sup (a_n+1)$. I want to show that $0<\lim \inf{a_n \over b_n}\le\lim \sup{a_n \over b_n}<\infty$ so as to conclude that $a_n$ and $b_n$ converge or diverge simultaneously. I can't however, find $\lim \sup$\ $\lim \inf$. A a matter of fact, for months I have known $\lim \sup$ and $\lim \inf$ but have never understood it for real, despite sitting for hours trying to understand. I would appreciate your help.


1 Answer 1


If $\sum\frac{a_n}{a_n+1}$ converges, we must have $\frac{a_n}{a_n+1}\to 0$. And for this we must have $a_n\to 0$.

Then I think you can use the limit comparison test: $\frac{a_n}{\left(\frac{a_n}{a_n+1}\right)}=a_n+1\to 1$

  • $\begingroup$ Thank you. One last thing: even though it is obvious that $a_n \to 0$, how is it actually showed? If $a_n\to L$ where $L\in\Bbb{R}$ or $L=\pm \infty$ that's trivial, but what if $a_n$ diverge, such that there is no limit, how do I contradict? $\endgroup$ Feb 26, 2015 at 10:15
  • $\begingroup$ $a_n=\frac{1}{1-\frac{a_n}{a_n+1}}-1$ $\endgroup$
    – paw88789
    Feb 26, 2015 at 10:25
  • $\begingroup$ How did you find it? There is no equality to work with. And by the way, can I say that ${a_n\over a_n+1}=f(n)\to 0$, and then conclude that $a_n=f(n)\cdot (a_n+1)\to 0$? $\endgroup$ Feb 26, 2015 at 10:27
  • $\begingroup$ Oh no, $\lim {a_n +1}$ is not defined. $\endgroup$ Feb 26, 2015 at 10:29
  • $\begingroup$ It doesn't need to be--in the expression on the right hand side of my previous comment, we just use the fact that $\frac{a_n}{a_n+1}\to 0$ (that's why I wrote $a_n$ in this manner). $\endgroup$
    – paw88789
    Feb 26, 2015 at 10:34

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