For more than one series of any search function known not seem to find the sum of this series: $$\sum_{n=0}^\infty{\frac{n}{3^n}}$$ I've found that converges with the quotient criterion.
Could you give me some suggestions for finding the sum of this series? Thanks!
Since $0<a= \frac{1}{3}<1$ the function $a^n$ converges uniformly, so you can interchange differentiation/integration and summation. Rewrite the expression as $$ a\sum_{k=0}^{\infty}\frac{da^k}{da}=a \cdot \frac{d}{da}\sum_{k=0}^{\infty}a^k $$ Can you handle from here?
Let $T=\sum_{n=0}^\infty\frac1{3^n}$, so $3T=3+\sum_{n=1}^\infty\frac1{3^{n-1}}=3+T$, and $T=3/2$.
Now let $S=\sum_{n=0}^\infty\frac n{3^n}$. So $$S/3=\sum_{n=0}^\infty\frac{n+1-1}{3^{n+1}}=\sum_{n=0}^\infty\frac{n+1}{3^{n+1}}-\sum_{n=0}^\infty\frac{1}{3^{n+1}}=(S-0)-(T-1),$$ and $$ S=\frac34. $$
(If needed, the manipulations with infinite series can be easily formalized by replacing them with manipulations with partial sums, and taking limits.)
