Finding the sum of a series with an n term in the numerator Sum the series:
$$\sum_{n=1}^\infty\frac{2n}{7^{2n-1}}$$
I know it converges, but it's not a geometric series nor is it power/telescoping/alternating. I think having the n term in the numerator makes it difficult to solve.
I took calculus BC a number of years ago and I don't think I remember learning how to do this. Any help would be greatly appreciated!

Consider the geometric sum $\sum\limits_{n=0}^{N-1}a^n=\frac{1-a^N}{1-a}$. Then,
$$\sum\limits_{n=1}^{N-1}na^{n-1}=\sum\limits_{n=0}^{N-1}na^{n-1}=\sum\limits_{n=0}^{N-1}\frac{\mathrm{d}}{\mathrm{d}a}a^n=\frac{\mathrm{d}}{\mathrm{d}a}\sum\limits_{n=0}^{N-1}a^n=\frac{\mathrm{d}}{\mathrm{d}a}\left(\frac{1-a^N}{1-a}\right)$$
Evaluating the derivative, we get
$$\frac{\mathrm{d}}{\mathrm{d}a}\left(\frac{1-a^N}{1-a}\right)=\frac{\left(-Na^{N-1}\right)\left(1-a\right)-\left(1-a^N\right)(-1)}{\left(1-a\right)^2}=\frac{1-a^N-Na^{N-1}\left(1-a\right)}{\left(1-a\right)^2}$$
Thus,
$$\sum\limits_{n=1}^{N-1}na^{n-1}=\frac{1-a^N-Na^{N-1}\left(1-a\right)}{\left(1-a\right)^2}$$
As $N\rightarrow\infty$, we get
$$\sum\limits_{n=1}^\infty na^{n-1}=\frac{1}{\left(1-a\right)^2}$$ for $|a|<1$. Now consider the series given.
$$\sum\limits_{n=1}^\infty\frac{2n}{7^{2n-1}}=\frac{2}{7}\sum\limits_{n=1}^\infty\frac{n}{7^{2\left(n-1\right)}}=\frac{2}{7}\sum\limits_{n=1}^\infty n\left(\frac{1}{49}\right)^{n-1}=\frac{2}{7}\times\frac{1}{\left(1-\frac{1}{49}\right)^2}=\frac{343}{1152}$$
So,
$$\sum\limits_{n=1}^\infty\frac{2n}{7^{2n-1}}=\frac{343}{1152}$$
This technique is widely used to evaluate sums and integrals in physics.
