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!

For your specific problem, rewrite $$\sum_{n=1}^\infty\frac{2n}{7^{2n-1}}=2 \times 7\sum_{n=1}^\infty\frac{n}{7^{2n}}=14\sum_{n=1}^\infty\frac{n}{49^{n}}$$ Now, consider 
$$\sum_{n=1}^\infty n x^n=x\sum_{n=1}^\infty n x^{n-1}=x\left(\sum_{n=1}^\infty  x^{n}\right)'$$
Finish and, when done, make $x=\frac 1 {49}$
