Question

If x > 0, then find :
$\sum\limits_{n\to1} ^\infty\left(\dfrac{x}{x+1}\right)^{n-1}$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given series is

$\rm :\longmapsto\:\sum\limits_{n\to1} ^\infty\left(\dfrac{x}{x+1}\right)^{n-1}$

Let first simplify the summation by substituting n = 1, 2, 3, --

So, we get

$\rm = \: 1 + \bigg[\dfrac{x}{x + 1} \bigg] + {\bigg[\dfrac{x}{x + 1} \bigg]}^{2} + {\bigg[\dfrac{x}{x + 1} \bigg]}^{3} + - -$

Now, its given that,

$\red{\rm :\longmapsto\:x > 0}$

$\red{\rm :\longmapsto\:x + 1 > x}$

$\red{\rm \implies\:\dfrac{x}{x + 1} \: < \: 1 \: }$

Further more,

$\rm = \: 1 + \bigg[\dfrac{x}{x + 1} \bigg] + {\bigg[\dfrac{x}{x + 1} \bigg]}^{2} + {\bigg[\dfrac{x}{x + 1} \bigg]}^{3} + - -$

represents an infinite GP series, whose

$\rm :\longmapsto\:a \: = \: 1$

and

$\rm :\longmapsto\:r \: = \: \dfrac{x}{x + 1}$

We know,

The sum of infinite GP series whose First term is a and Common ratio r respectively, then

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_ \infty \:=\: \frac{a}{1 - r} \: \: provided \: that \: |r| < 1 }}}}}} \\ \end{gathered}$

Now,

We have

$\red{\rm :\longmapsto\:x > 0}$

$\red{\rm \implies\:\dfrac{x}{x + 1} > 0}$

and

$\red{\rm \implies\:\dfrac{x}{x + 1} < 1}$

So,

$\red{\rm \implies\:0 < \dfrac{x}{x + 1} < 1}$

So,

$\rm \: 1 + \bigg[\dfrac{x}{x + 1} \bigg] + {\bigg[\dfrac{x}{x + 1} \bigg]}^{2} + {\bigg[\dfrac{x}{x + 1} \bigg]}^{3} + - -$

$\rm = \: \dfrac{1}{1 - \dfrac{x}{x + 1} }$

$\rm = \: \dfrac{1}{\dfrac{x + 1 - x}{x + 1} }$

$\rm = \: \dfrac{1}{\dfrac{1}{x + 1} }$

$\rm = \: x + 1$

Hence,

$\rm \implies\:\boxed{ \tt{ \: \sum\limits_{n\to1} ^\infty\left(\dfrac{x}{x+1}\right)^{n-1} = x + 1 \: }}$

Additional Information :-

The general term of GP series

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\: {r}^{n \: - \: 1} }}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• r is the common ratio.

Sum of n terms of GP series

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\: \frac{a \: ( {r}^{n} \: - \: 1) }{r \: - \: 1} }}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of AP.

• a is the first term of the sequence.

• n is the no. of terms.

• r is the common ratio.