# How to calculate the limit of this sum with different methods? [duplicate]

It's a basic question , but what are the common methods to calculate limits like this one: $$\sum_{k=1}^\infty \frac{3k}{7^{k-1}}$$

• This is a duplicate of a duplicate of a duplicate. Commented Aug 19, 2015 at 16:35
• @columbus8myhw Correct, and I voted accordingly.
– user147263
Commented Aug 19, 2015 at 22:30

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,

$$\sum_{n=1}^\infty\frac{3n}{7^{n-1}}=\frac 3{(1-1/7)^2}.$$

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: $$\sum_{k\geq 1}\frac{3k}{7^{k-1}} = 21\cdot S = 21\cdot\frac{7}{36} = \color{red}{\frac{49}{12}}.$$

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}