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showing 1-1 of 1 (answers : 1, comments : 0),   sorted newest to the oldest
Question
How to simplify this sequence?
$x^{-1}+2x^{-2}+3x^{-3}+\cdots$

Also find the values of $x$ for which the summation would converge.

Ans: $\dfrac{x}{(x-1)^2}$
 
(0) (0) Comment Answer
2017-01-31 21:01:46 
Anwesa Roy
1 Answer
$s=x^{-1}+2x^{-2}+3x^{-3}+4x^{-4}+\cdots $

$sx^{-1}=x^{-2}+2x^{-3}+3x^{-4}+4x^{-5}+\cdots$

$s-sx^{-1}=x^{-1}+x^{-2}+x^{-3}+x^{-4}+\cdots\\
\Rightarrow s\left(1-\dfrac{1}{x}\right)=\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{1}{x^3}+\dfrac{1}{x^4}+\cdots\\
\Rightarrow s\left(1-\dfrac{1}{x}\right)=\dfrac{\dfrac{1}{x}}{1-\dfrac{1}{x}}\\
\Rightarrow s=\dfrac{\dfrac{1}{x}}{\left(1-\dfrac{1}{x}\right)\left(1-\dfrac{1}{x}\right)}\\
\Rightarrow \boxed{s=\dfrac{x}{(x-1)^2}}$

We have written it as a geometric progression and it can be seen that the series converges when
$\dfrac{1}{|x|}\lt 1$
 
(0) (0) Comment
2017-01-31 21:28:48 
jiju (Junior Maths Expert, careerbless.com)
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