Question

start with the geometric series $\sum_{n = 0}^{\infty}x^{n}$. (a) find the sum of the series $\sum_{n = 1}^{\infty}nx^{n - 1}, |x| < 1$. (b) find the sum of each of the following series. (i) $\sum_{n = 1}^{\infty}nx^{n}, |x| < 1$ (ii) $\sum_{n = 1}^{\infty}\frac{n}{7^{n}}$

Explanation:

Step1: Recall the formula for geometric series

The geometric series \(\sum_{n = 0}^{\infty}x^{n}=\frac{1}{1 - x}\), for \(|x|\lt1\).

Step2: Differentiate the geometric series term - by - term

Differentiate both sides of \(\sum_{n = 0}^{\infty}x^{n}=\frac{1}{1 - x}\) with respect to \(x\).
Using the power rule \((x^{n})^\prime=nx^{n - 1}\) and the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\) (where \(u = 1\), \(v=1 - x\), \(u^\prime=0\), \(v^\prime=- 1\)).
\(\sum_{n = 1}^{\infty}nx^{n - 1}=\frac{0\times(1 - x)-1\times(-1)}{(1 - x)^{2}}=\frac{1}{(1 - x)^{2}}\)

Step3: Solve part (b)(i)

Multiply the result of part (a) by \(x\).
If \(S=\sum_{n = 1}^{\infty}nx^{n-1}=\frac{1}{(1 - x)^{2}}\), then \(\sum_{n = 1}^{\infty}nx^{n}=x\sum_{n = 1}^{\infty}nx^{n - 1}\)
So \(\sum_{n = 1}^{\infty}nx^{n}=\frac{x}{(1 - x)^{2}}\), for \(|x|\lt1\)

Step4: Solve part (b)(ii)

Let \(x=\frac{1}{7}\) in the formula \(\sum_{n = 1}^{\infty}nx^{n}=\frac{x}{(1 - x)^{2}}\)
Substitute \(x = \frac{1}{7}\) into \(\frac{x}{(1 - x)^{2}}\), we have \(1-x=1-\frac{1}{7}=\frac{6}{7}\)
\(\sum_{n = 1}^{\infty}\frac{n}{7^{n}}=\frac{\frac{1}{7}}{(\frac{6}{7})^{2}}=\frac{\frac{1}{7}}{\frac{36}{49}}=\frac{7}{36}\)

Answer:

(a) \(\frac{1}{(1 - x)^{2}}\)

(b)(i) \(\frac{x}{(1 - x)^{2}}\)

(b)(ii) \(\frac{7}{36}\)