Do the following series converge if $a_n>0$ and $ \sum_{n=1}^{\infty}a_n$ diverges? Do the following series converge if $a_n>0$ and $\sum_{n=1}^{\infty}a_n$ diverges ?
a.) $\sum_{n=1}^{\infty}{a_n \over 1+ a_n}$
b.)$\sum_{n=1}^{\infty}{a_n\over 1+ a_n ^2}$
c.)$\sum_{n=1}^{\infty}{a_n\over 1+ na_n }$
d.)$\sum_{n=1}^{\infty}{a_n\over 1+ n^2a_n }$
What I thought was using Alembert criteria , but it just not seem to work. I also now if $a_n$ converges so does $a_n^2$. Ive tried everything to no avail, can anyone please help ?
 A: For (a),  if it is convergent then $a_n/(1+a_n)\to 0$, then $a_n \to 0$, then using 
$$a_n \sim \frac{a_n}{1 + a_n}$$
when $a_n\to 0$, then we know $\sum a_n$ must converge, contradiction with $\sum a_n$ diverges.
for (b), it is unknown, 
when $a_n = 1/n$, $\frac{a_n}{1+a_n^2} = \frac{n}{1+n^2} \sim \frac{1}{n}$, thus diverges.
when $a_n = n^2$, $\frac{a_n}{1 + a_n^2} = \frac{n^2}{1 + n^4}\sim \frac{1}{n^2}$, converges.
for (c), it is unknown too.
when $a_n = 1/n$, $\frac{a_n}{1 + na_n} = \frac{a_n}{2}$, thus diverges.
think about 
$a_n = n$ if $n = k^2$ for some $k\in N$.
$a_n = \frac{1}{n^2}$ for other $n$ which is not a square.
In this case,
$\sum \frac{a_n}{1 + na_n} = \sum_{n\in\{k^2, k\in N\}} \frac{n}{1 + n^2} + \sum_{others} \frac{1}{n^2 + n}$,
first part is equivalent to $\sum_k \frac{k^2}{1+ k^4}$ converges.
second part is equivalent to $\sum_n \frac{1}{n^2 +n}$ converges.
The last one, each term is less than $1/n^2$, thus converges.
A: b) It might converge, it might diverge: Look at $a_n = n^2, a_n = 1/n.$
c) It might converge, it might diverge: $a_n = 1/n \implies \sum a_n/(1+na_n) = \infty.$ For an example where $\sum a_n = \infty, \sum a_n/(1+na_n) < \infty,$ define $$\begin{cases} a_{2^m} = 1, m = 1,2,\dots \\a_n = 1/n^2, n \not \in \{2^m: m =1,2,\dots \}\end{cases}$$
