How to calculate the limit of this sum with different methods? It's a basic question , but what are the common methods to calculate limits like this one:
$$\sum_{k=1}^\infty \frac{3k}{7^{k-1}}$$
 A: Using the fact that 
$$\sum_{n=0}^\infty x^n=\frac 1{1-x},\quad |x|<1,$$
differentiate both sides to get
$$\sum_{n=1}^\infty nx^{n-1}=\frac 1{(1-x)^2},\quad |x|<1.$$
Now, 
$$\sum_{n=1}^\infty\frac{3n}{7^{n-1}}=\frac 3{(1-1/7)^2}.$$
A: Let $S=\sum_{k\geq 1}\frac{k}{7^k}$. Then:
$$ 6S = 7S-S = \sum_{k\geq 1}\frac{k}{7^{k-1}}-\sum_{k\geq 1}\frac{k}{7^k} = \sum_{k\geq 0}\frac{k+1}{7^k}-\sum_{k\geq 1}\frac{k}{7^k}=1+\sum_{k\geq 1}\frac{1}{7^k}=\frac{7}{6}$$
hence:
$$ \sum_{k\geq 1}\frac{3k}{7^{k-1}} = 21\cdot S = 21\cdot\frac{7}{36} = \color{red}{\frac{49}{12}}. $$
A: Yet another way is to recognize that the single sum can be written as a double sum.  To that end, we have
$$\begin{align}
\sum_{k=1}^{\infty}\frac{3k}{7^{k-1}}&=21\sum_{k=1}^{\infty}\frac{k}{7^k}\\\\
&=21\sum_{k=1}^{\infty}\frac{1}{7^k}\sum_{\ell=1}^{k}\,1\\\\
&=21\sum_{\ell=1}^{\infty}\sum_{k=\ell}^{\infty}\frac{1}{7^k}\\\\
&=21\sum_{\ell=1}^{\infty}\frac{(1/7)^{\ell}}{1-(1/7)}\\\\
&=(21)(7/6)\frac{1/7}{1-(1/7)}\\\\
&=\frac{49}{12}
\end{align}$$
