Question

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

Answer 1

Step-by-step explanation:

Given:$\sum^{ \infty}_{n = 1} \left( \dfrac{x}{x + 1} \right)^{n - 1}$

To find: Evaluate the summation if x>0

Solution:

Step 1 :Expand the summation

$\sum^{ \infty}_{n = 1} \left( \dfrac{x}{x + 1} \right)^{n - 1}$

$= \left( { \frac{x}{x + 1} }\right)^{0} + \left( { \frac{x}{x + 1} }\right)^{1} +\left( { \frac{x}{x + 1} }\right)^{2} + \left( { \frac{x}{x + 1} }\right)^{3} + ... ∞$

Simplify

$= 1 + \frac{x}{x + 1} + \left( { \frac{x}{x + 1} }\right)^{2} + \left( { \frac{x}{x + 1} }\right)^{3} + ...∞$

Step 2: Identify the series

It is an infinite GP

its first term is, a= 1

it's common ratio is,r= x/x+1

Step 3: Apply sum of infinite GP

$S_∞=\frac{a}{1 - r}$

if r<1

Apply the sum

$= \frac{1}{1 - \frac{x}{x + 1}} \\ \\ = \frac{1}{ \frac{ \cancel x + 1 - \cancel x}{x + 1} } \\ \\ = \frac{1}{ \frac{1}{x + 1} } \\ \\ = x + 1$

Final answer:

$\bold{\red{\sum^{ \infty}_{n = 1} \left( \dfrac{x}{x + 1} \right)^{n - 1} =x+1}}$

Hope it helps you.

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