I have the following series: ∑ n = 0 ∞ n ( 2 z − 1 ) n

Question

I have the following series:      ∑          n      =      0        ∞    n  (  2  z  −  1      )    n

Punktatsp · Accepted Answer

There is no need to do any transformation. The radius of convergence of

\sum_{n = 0}^{\infty} n z^{n}

is just

\underset{n \to \infty}{lim sup} \sqrt[n]{n} = 1

Therefore, when

| 2 z - 1 | < 1

, the series

\sum_{n = 0}^{\infty} n (2 z - 1)^{n}

converge. And if

| 2 z - 1 | \geq 1

, the term

| n (2 z - 1)^{n} | \to \infty \neq 0

, so the series diverge.

polivijuye · Accepted Answer

To begin with, I would start with noticing that

\begin{aligned} \sum_{n = 0}^{\infty} n (2 z - 1)^{n} & = \sum_{n = 1}^{\infty} n (2 z - 1)^{n} \\ = (2 z - 1) \sum_{n = 1}^{\infty} n (2 z - 1)^{n - 1} \\ = (2 z - 1) \sum_{n = 0}^{\infty} (n + 1) (2 z - 1)^{n} \end{aligned}

Moreover, we also have the geometric series:

\begin{aligned} \frac{1}{1 - w} = 1 + w + w^{2} + \dots = \sum_{n = 0}^{\infty} w^{n} \end{aligned}

which converges whenever

| w | < 1

. Due to its properties, we can derive both sides in order to get:

\begin{aligned} {(\frac{1}{1 - w})}^{2} = \sum_{n = 1}^{\infty} n w^{n - 1} = \sum_{n = 0}^{\infty} (n + 1) w^{n} \end{aligned}

Comparing both results, we conclude that

\begin{aligned} \sum_{n = 0}^{\infty} n (2 z - 1)^{n} & = (2 z - 1) \sum_{n = 0}^{\infty} (n + 1) (2 z - 1)^{n} = \frac{2 z - 1}{4 (1 - z)^{2}} \end{aligned}

which converges whenever

| 2 z - 1 | < 1

, and we are done.

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