Showing that $\sum\limits_{n=1}^{\infty} (a_1+2a_2+…+na_n)/n(n+1) = \sum\limits_{i=1}^n a_n$

Let $\sum\limits_{n=1}^{\infty} a_n$ a series of positive terms convergent. Show that $\sum\limits_{n=1}^{\infty} \frac{a_1+2a_2+...+na_n}{n(n+1)}$ converges to the same value of $\sum\limits_{i=1}^n a_n$

I think I can use the Cauchy condensation test but I would like to know if there is an easier way.

• Dear jggarita, I see you are new to math.stackexchange, so let me explain you some things. On math.stackexchange, we expect that your share your thoughts about the problem and show what you have already done, when it concerns a self-study question. If you don't do that, some users might downvote your question for not giving this information (to indicate that the question can still be improved). – Pedro Apr 16 '15 at 22:09

If we set $A_n=\sum_{k=1}^{n}ka_k$, since $\sum_{n=a}^b\frac{1}{n(n+1)}=\frac{1}{a}-\frac{1}{b+1}$ (it is a telescopic sum) we have: $$\sum_{n=1}^{N}\frac{A_n}{n(n+1)}=\sum_{n=1}^{N}a_n\sum_{k=n}^{N}\frac{n}{k(k+1)}=\sum_{n=1}^{N}a_n\left(1-\frac{n}{N+1}\right).$$ To prove our claim it is sufficient to notice that the last sum is the average of the first $N+1$ partial sums of $\sum_{n\geq 1}a_n$ (the order-zero partial sum being zero): provided that the last series is converging, Césaro theorem now gives: $$\sum_{n\geq 1}\frac{A_n}{n(n+1)}=\sum_{n\geq 1}a_n$$ as wanted.
• I think there is an elementary argument after your $\sum_{n=1}^N a_n (1 - n/N)$ formula that only takes a few steps to show it converges to $\sum_n a_n$,, instead of using Cesaro, but I agree you have now basically made the proof "complete" – user2566092 Apr 17 '15 at 15:25