Convergence of $\sum\limits_n\frac{1}{a_n+1}$ and $\sum\limits_n\frac{a_n}{a_n+1}$ when $\sum\limits_na_n$ converges This is from an MCQ contest.

Let  $\sum\limits_{n\geq 1}a_n$ be a convergent series of positive terms. Which of the following hold? 
  
  
*
  
*$1]$ $\sum\limits_{n\geq 1}\dfrac{1}{a_n+1}$ and $\sum\limits_{n\geq 1}\dfrac{a_n}{a_n+1}$ are convergent.
  
*$2]$ $\sum\limits_{n\geq 1}\dfrac{1}{a_n+1}$ and $\sum\limits_{n\geq 1}\dfrac{a_n}{a_n+1}$ are divergent. 
  
*$3]$ $\sum\limits_{n\geq 1}\dfrac{1}{a_n+1}$ is Divergent and  $\sum\limits_{n\geq 1}\dfrac{a_n}{a_n+1}$ is convergent 
  
*$4]$ none of the previous statements is correct
  

i come up with this conter example :


*

*$1]$ $\sum_{n\geq 1} \dfrac{1}{n^2}$ convergent  so let 's verify:
$\sum\limits_{n\geq 1}\dfrac{1}{a_n+1}=\sum\limits_{n\geq 1}\dfrac{n^2}{1+n^{2}}$


or $\lim_{n\to +\infty }\dfrac{n^2}{1+n^{2}}=1 \neq 0$ then $\sum\limits_{n\geq 1}\dfrac{1}{a_n+1}$ divergent thus $1]$ False i don't need to verify $\sum\limits_{n\geq 1}\dfrac{a_n}{a_n+1}$ because there is  'and' in the statement $1]$ 


*

*$2]$ in this case i need to check the nature of $\sum\limits_{n\geq 1}\dfrac{a_n}{a_n+1}$ 


$\sum\limits_{n\geq 1}\dfrac{a_n}{a_n+1}=\sum\limits_{n\geq 1}\dfrac{1}{1+n^2}$
i don't know to calculate i use just Wolframe and it's convergent


*

*$3]$ i can't say that statement is true just becuase one example work i need to prove it but i don't now how

*Is my proof correct

*Is there any kind of reasoning one may use during a contest, when a quick answer is needed

 A: From your solution of (1), I'll assume that $\sum a_n$ is convergent. If so, your argument for (1) is correct. For (2) and (3) (and also (1)) note that


*

*$\displaystyle\frac{a_n}{a_n+1}<a_n$ so $\sum\dfrac{a_n}{a_n+1}$ is convergent.

*$\dfrac{1}{a_n+1}\to 1$ as $n\to \infty$ so $\sum\frac{1}{a_n+1}$ is divergent.


I hope that also answers your last question. Though honestly I'd say there's no magic recipe in handling convergence of a series.
A: Much of the analysis in the OP was fine.  To add some thoughts, recall that if a series converges, then it's terms must approach zero as $n\to 0$.  
Therefore, if $\sum_{n=1}^{\infty}a_n$ converges, then $\lim_{n\to \infty}a_n=0$.  This implies that $\lim_{n\to \infty}\frac{1}{1+a_n}=1$, which in turn implies that $\sum_{n=1}^{\infty}\frac{1}{1+a_n}$ diverges.
Now, since $a_n\to 0$, then there exists a number $N$ so that $a_n<1$ whenever $n>N$.  Then, for such an $N$, $\frac{a_n}{1+a_n}<\frac12 a_n$.  Inasmuch as the series $\sum_{n=1}^{\infty}\frac12 a_n$ converges, then so does the series $\sum_{n=1}^{\infty}\frac{a_n}{1+a_n}$.
The answer is, therefore, number $3$.
