\(\sum\limits_{n=0}^{\infty}nx^{n-1},\,|x|<1\).

The main purpose of this question is to find the sum of the series $\sum\limits_{n=0}^{\infty}nx^{n-1}$ starting with $\sum\limits_{n=0}^{\infty}x^n$.

The concept of sequence and series is one of the most fundamental concepts in arithmetic. A sequence can be referred to as a detailed list of elements with or without repetition, while a series is a sum of all the elements of a sequence. Some of the very common types of series include arithmetic series, geometric series, and harmonic series.

Suppose that $\{a_k\}=1,2,\cdots$ is a sequence with each successive term calculated by adding a constant $d$ to the preceding term. In this series, the sum of the first $n$ terms is given by $S_n=\sum\limits_{k=1}^{n}a_k$ where $a_k=a_1+(k-1)d$.

The sum of terms in a geometric sequence is regarded as the geometric series and has the following form:

$a+ar+ar^2+ar^3+\cdots$

where $r$ is said to be the common ratio.

Mathematically, a geometric series $\sum\limits_{k}a_k$ is one in which the ratio of two successive terms $\dfrac{a_{k+1}}{a_{k}}$ is a constant function of the summation index $k$.

The series $\sum\limits_{n=1}^{\infty}\dfrac{1}{n}$ is said to be harmonic series. This series can be regarded as the series of rational numbers having integers in the denominator (in an increasing manner) and a one in the numerator. Harmonic series can be used for comparisons due to their divergent nature.

## Expert Answer

The given geometric series is:

$\sum\limits_{n=0}^{\infty}x^n=1+x+x^2+x^3+\cdots$

The closed form of this series is:

$\sum\limits_{n=0}^{\infty}x^n=\dfrac{1}{1-x}$

Since, $\sum\limits_{n=0}^{\infty}nx^{n-1}=1+2x+3x^2+4x^3+\cdots$ (1)

$=(1+x+x^2+x^3+\cdots)+(x+2x^2+3x^3+4x^4+\cdots)$

As $1+x+x^2+x^3+\cdots=\dfrac{1}{1-x}$, therefore we get:

$\sum\limits_{n=0}^{\infty}nx^{n-1}=\dfrac{1}{1-x}+x(1+2x+3x^2+4x^3+\cdots)$

And from (1):

$\sum\limits_{n=0}^{\infty}nx^{n-1}=\dfrac{1}{1-x}+x\sum\limits_{n=0}^{\infty}nx^{n-1}$

$\sum\limits_{n=0}^{\infty}nx^{n-1}-x\sum\limits_{n=0}^{\infty}nx^{n-1}=\dfrac{1}{1-x}$

$(1-x)\sum\limits_{n=0}^{\infty}nx^{n-1}=\dfrac{1}{1-x}$

$\sum\limits_{n=0}^{\infty}nx^{n-1}=\dfrac{1}{(1-x)^2}$

## Example 1

Determine the sum of infinite geometric sequence starting at $a_1$ and has $n^{th}$ term $a_n=2\times 13^{1-n}$.

### Solution

For $n=1$, $a_1=2\times 13^{1-1}$

$=2\times 13^0$

$=2\times 1$

$=2$

For $n=2$, $a_2=2\times 13^{1-2}$

$=2\times 13^{-1}$

$=\dfrac{2}{13}$

Now, $r=\dfrac{2}{13}\div 2=\dfrac{1}{13}$

Since $|r|<1$, so the given series is convergent with sum:

$S_{\infty}=\dfrac{a_1}{1-r}$

Here, $a_1=2$ and $r=\dfrac{1}{13}$.

Therefore, $S_{\infty}=\dfrac{2}{1-\dfrac{1}{13}}$

$S_{\infty}=\dfrac{26}{12}=\dfrac{13}{6}$

## Example 2

Given the infinite geometric series:

$1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\cdots$, find its sum.

### Solution

First find the common ratio $r$:

$r=\dfrac{\dfrac{1}{3}}{1}=\dfrac{1}{3}$

Since the common ratio $|r|<1$ therefore, the sum of infinite geometric series is given by:

$S_{\infty}=\dfrac{a_1}{1-r}$

where $a_1$ is the first term.

$S_{\infty}=\dfrac{1}{1-\dfrac{1}{3}}=\dfrac{3}{2}$

## Example 3

Given the infinite geometric series:

$\dfrac{12}{1}+\dfrac{12}{2}+\dfrac{12}{3}+\cdots$, find its sum.

### Solution

First find the common ratio $r$:

$r=\dfrac{\dfrac{12}{2}}{\dfrac{12}{1}}=\dfrac{12}{2}\times \dfrac{1}{12}=\dfrac{1}{2}$

Since the common ratio $|r|<1$ therefore, the sum of infinite geometric series is given by:

$S_{\infty}=\dfrac{a_1}{1-r}$

where $a_1=\dfrac{1}{2}$ is the first term.

$S_{\infty}=\dfrac{\dfrac{12}{1}}{1-\dfrac{1}{2}}=24$