How to prove this series $\sum_{n=1}^{\infty}\dfrac{a_{n}}{(n+1)a_{n+1}}$ diverges Question:

Assume that $a_{n}>0,n\in N^{+}$, and that  $$\sum_{n=1}^{\infty}a_{n}$$ is convergent. Show that
  $$\sum_{n=1}^{\infty}\dfrac{a_{n}}{(n+1)a_{n+1}}$$ is divergent?

My idea: since
$\sum_{n=1}^{\infty}a_{n}$ converges, then there exists $M>0$ such
$$a_{1}+a_{2}+\cdots+a_{n}<M$$
then I can't continue. Thank you.
 A: This is essentially the same as Erick Wong's lovely solution to a similar problem.
Set $\displaystyle b_n=\frac{a_n}{(n+1)a_{n+1}}$.  Since $a_n\to 0$ and $a_n>0$ there is some decreasing subsequence $a_{n_1}>a_{n_2}>a_{n_3}>\cdots$.  By passing to a subsubsequence if necessary, we may assume without loss that $n_{i+1}\ge 2n_i+1$ for each $i$.  Set $d_i=n_{i+1}-n_i$.  The strategy now will be to bound $\displaystyle \sum_{n_i\le k<n_{i+1}}b_k$ below by $\frac{1}{2}$ using AM-GM; this proves that the desired series diverges.  We have
$$\prod_{n_i\le k<n_{i+1}}b_k>\prod_{n_i\le k<n_{i+1}}\frac{1}{n_{i+1}+1}\frac{a_k}{a_{k+1}}=\left(\frac{1}{n_{i+1}+1}\right)^{d_i}\frac{a_{n_i}}{a_{n_{i+1}}}>\left(\frac{1}{n_{i+1}+1}\right)^{d_i}$$
Now, by AM-GM, we have $$\sum_{n_i\le k<n_{i+1}}b_k\ge d_i\left(\frac{1}{n_{i+1}+1}\right)\ge \frac{1}{2}$$
Where the last inequality follows from rearranging $\displaystyle \frac{n_{i+1}}{2}\ge n_i+\frac{1}{2}$ as $$d_i=n_{i+1}-n_i\ge \frac{1}{2}(n_{i+1}+1)$$
A: Longtime ago,I post my methods
proof:Assmue that: $\displaystyle\sum_{n=1}^{\infty}\dfrac{a_{n}}{(n+1)a_{n+1}}$,then for any $p\in N^{+}$,have 
$$\sum_{k=n}^{n+p-1}\dfrac{a_{k}}{(k+1)a_{k+1}}<\dfrac{1}{4}\Longrightarrow\sum_{k=n}^{n+p-1}\dfrac{a_{k}}{(n+p)a_{k+1}}<\dfrac{1}{4}$$
then 
$$\dfrac{1}{p}\sum_{k=n}^{n+p-1}\dfrac{a_{k}}{a_{k+1}}<\dfrac{1}{4}\cdot\dfrac{n+p}{p}<\dfrac{1}{2}(p>n\mbox{时})$$
other hand,By AM-GM
$$\dfrac{1}{p}\sum_{k=n}^{n+p-1}\dfrac{a_{k}}{a_{k+1}}=\dfrac{1}{p}\left(\dfrac{a_{n}}{a_{n+1}}+\dfrac{a_{n+1}}{a_{n+2}}+\cdots+\dfrac{a_{n+p-1}}{a_{n+p}}\right)$$
so $$\sqrt[n]{\dfrac{a_{n}}{a_{n+p}}}<\dfrac{1}{2}\Longrightarrow a_{n+p}>2^na_{n}$$
contradiction
