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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.
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.
