Finding the sum of a series with an n term in the numerator

Question

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!

Essentially, you will find your answer here (look at the first answer, where the partial sum is computed) https://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-inftyn1xn — Suzet, Jul 13 '18 at 03:48
You misunderstand me. I mean it's the power series $\sum_{n=1}^{\infty}2nx^{2n-1}$ evaluated at $x=\frac17$ — saulspatz, Jul 13 '18 at 04:14

score 2 · Answer 1 · answered Jul 13 '18 at 04:14

2

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 $x=\frac 1 {49}$

answered Jul 13 '18 at 04:14

Claude Leibovici

260,315

You need to show that the derivative can be taken outside the infinite sum. To do this you need to first evaluate the partial sum (where taking the derivative works as it is a finite sum) and then take the limit. I've done this in my answer below. – Chetan Vuppulury Jul 13 '18 at 04:31
@ChetanVuppulury: Any power series can be differentiated or integrated in the interior of region of convergence. – Paramanand Singh Jul 13 '18 at 04:57
Oh. I didn't know that. Can you give me a reference? – Chetan Vuppulury Jul 13 '18 at 05:20

score 1 · Answer 2 · answered Jul 13 '18 at 04:29

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.

mengdie1982 · Answer 3 · 2018-07-13T05:44:23.673

$$\sum_{n=1}^\infty\frac{2n}{7^{2n-1}}=2 \times 7\sum_{n=1}^\infty\frac{n}{7^{2n}}=14\sum_{n=1}^\infty\frac{n}{49^{n}}$$

$\sum\limits_{n=1}^\infty\dfrac{n}{49^{n}}$ 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, $$14\sum_{n=1}^\infty\frac{n}{49^{n}}=14 \cdot \frac{49}{48^2}=\frac{343}{1152}.$$

Finding the sum of a series with an n term in the numerator

3 Answers3