Sum of the series $\sum\limits_{n=1}^{\infty }\frac{n}{3^n}$ I want to calculate the sum: $$\sum _{n=1}^{\infty }\:\frac{n}{3^n}\:$$
so $:\:\sum_{n=1}^{\infty}\:nx^n;\:x=\frac{1}{3}\:$ 
$$=x\sum_{n=1}^{\infty}\:nx^{n-1}=x\sum_{n=1}^{\infty}\:n\:\left(\int\left(x^{n-1}\right)dx\right)'=x\sum_{n=1}^{\infty}\:\left(x^n\right)'
$$
now from here I would continue: $x\:\left(\frac{x}{1-x}\right)'=\frac{x}{^{\left(1-x\right)^2}}\:\left(\frac{1}{3}\right)=\frac{3}{4}$
In the answer that I saw, there is another step, from which we get the same result, but I don't understand why it is correct to do so:
$$
x\sum_{n=1}^{\infty}\:\left(x^n\right)'=x\sum_{n=0}^{\infty} ({x^{n}})' =x\cdot \left(\frac{1}{1-x}\right)'=\frac{x}{^{\left(1-x\right)^2}}
$$
Is this just a spelling mistake ?
 A: Another approach is the following: the series is absolutely convergent since $3^n\geq n^3$ for any $n\geq 3$.
If we set $S=\sum_{n\geq 1}\frac{n}{3^n}$, we have:
$$2S = 3S-S = \sum_{n\geq 1}\frac{3n}{3^n}-\sum_{n\geq 1}\frac{n}{3^n} = \sum_{n\geq 0}\frac{n+1}{3^n}-\sum_{n\geq 1}\frac{n}{3^n} = 1+\sum_{n\geq 1}\frac{1}{3^n},$$
hence $2S=1+\frac{1}{2}$ leads to $S=\color{red}{\frac{3}{4}}$ as wanted.
A: As jschnei points out in the comments, the step follows because the derivative of any constant is always zero (in other words, the answer you saw added zero in a convenient way).
A: Although this does not address the specific question, I thought it might be instructive to present another approach for solving a problem of this nature.  So, here we go
Let $S=\sum_{n=1}^\infty nx^n.\,\,$   Note that we could also write the sum $S$ as $S=\sum_{n=0}^\infty nx^n,\,\,$ since the first term $nx^n=0$ for $n=0$. We will use the former designation in that which follows.
Observing that we can write $n$ as $n=\sum_{m=1}^n (1)$ (or $n=\sum_{m=0}^{n-1}(1)$), the series of interest $S$ can be written therefore
$$\begin{align}
S&=\sum_{n=1}^\infty \left(\sum_{m=1}^n (1)\right)x^n\\\\
&=\sum_{n=1}^\infty \sum_{m=1}^n (1)x^n\
\end{align}$$
Now, simply changing the order of summation yields
$$\begin{align}
S&=\sum_{m=1}^\infty (1)\left(\sum_{n=m}^\infty x^n\right)\\\\
&=\sum_{m=1}^\infty (1)\left(\frac{x^m}{1-x}\right)\\\\
&=\frac{x}{(1-x)^2}
\end{align}$$
which recovers the result obtained through the well-known methodology of differentiation under the summation sign.
