Positive series problem: $\sum\limits_{n\geq1}a_n=+\infty$ implies $\sum_{n\geq1}\frac{a_n}{1+a_n}=+\infty$ Let $\sum\limits_{n\geq1}a_n$ be a positive series, and $\sum\limits_{n\geq1}a_n=+\infty$, prove that: $$\sum_{n\geq1}\frac{a_n}{1+a_n}=+\infty.$$
 A: I would argue by cases.
Case 1. $a_n\ge1$ for infinitely many $n\in\mathbb N$.
In this case, for each such $n$ we have $\frac{a_n}{1+a_n}=1-\frac{1}{1+a_n}\ge1-\frac12$, from which the claim easily follows.
Case 2. $a_n\ge1$ for only finitely many $n\in\mathbb N$.
In this case for every other $n$ we have $a_n&lt1$ and thus $\frac{a_n}{1+a_n}\ge\frac{a_n}{1+1}=\frac{a_n}2$. Since finitely many terms can't affect the convergence/divergence of a series, this will also diverge. (Since $\sum\limits_{n=1}^\infty\frac{a_n}{2}$ does.)
A: Alternatively, split the problem up in cases: 
1, If there is a natural $N\in\mathbb{N}$ s.t $a_n\leq{1}$ for $n\geq N$, what can you conclude?
2, If (1) is not true, for every natural $N$ we can find a $n\geq{N}$ s.t $a_n>1.$ Now passing to a subsequence and comparing with a series with each term equal to a constant (more precisely $1/2$ or lower), the result follows. 
A: Proving the contrapositive statement seems cleaner to me. Suppose $\sum{a_n\over 1+{a_n}}$ converges. Then ${a_n\over 1+{a_n}}\rightarrow 0$. This implies that ${a_n}$ is eventually less than one, so ${a_n\over2}\le {a_n\over a_n+1}$ for $n$ sufficiently large. The comparision test then shows that $\sum a_n$ converges.
A: You can suppose that $(a_n) \rightarrow 0$ (if not the problem is trivial). Then what can you say asymptotically about $\frac{a_n}{1+a_n}$ ?
A: We can divide into  cases:


*

*If a(n) has limit zero : It is lower than 1 for all n bigger than n0, then we can compare with a(n)/2 which is lower than  a(n)/(1+a(n)).

*If a(n) has limit different to zero , also a(n)/1+a(n) and then the series diverges

*If a(n) is not bounded it ha a subsequence that converges to infinite,
then a(n)/1+a(n) converges to 1 then the series diverges to infinite.

*If a(n) is bounded , we can take a subsequence that is convergent.
If it does not converges to zero also the sequence a(n)/1+a(n).
If all subsequences converge to zero ,then also a(n) and we can apply 1.
A: Suppose $\sum{a_n\over 1+{a_n}}$ converges. Then $\frac{a_n}{1+a_n}\to 0$.
It is easy to see that:
$$\lim_{n\to\infty}a_n=0\iff \lim_{n\to\infty}\frac{a_n}{1+a_n}=0.$$
(let $b_n=\frac{a_n}{1+a_n}$, then $a_n=\frac{b_n}{1-b_n}$
)
So we have
$$\lim_{n\to\infty}\frac{\frac{a_n}{1+a_n}}{a_n}=1,$$
by comparision test of positive series,
the series $\sum a_n$ converges, it is a contradiction.
