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!
 A: 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 $x=\frac 1 {49}$
A: 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.
A: $$\sum_{n=1}^\infty\frac{2n}{7^{2n-1}}=2 \times 7\sum_{n=1}^\infty\frac{n}{7^{2n}}=14\sum_{n=1}^\infty\frac{n}{49^{n}}$$
$\sum\limits_{n=1}^\infty\dfrac{n}{49^{n}}$ is convergent, which would be easily proven by  Cauchy radical test method. Hence, we may denote $$S=\sum_{n=1}^\infty\dfrac{n}{49^{n}}=\frac{1}{49}+\frac{2}{49^2}+\frac{3}{49^3}+\cdots. \tag1$$
Thus, $$49\cdot S=1+\frac{2}{49}+\frac{3}{49^2}+\frac{4}{49^3}+\cdots.\tag2$$
By $(2)-(1)$, $$48S=1+\frac{1}{49}+\frac{1}{49^2}+\frac{1}{49^3}+\cdots=\frac{1}{1-\dfrac{1}{49}}=\frac{49}{48}.$$
Thus, $$S=\frac{49}{48^2}.$$
As a result, $$14\sum_{n=1}^\infty\frac{n}{49^{n}}=14 \cdot \frac{49}{48^2}=\frac{343}{1152}.$$
