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 November 14th, 2017, 08:31 AM #1 Member   Joined: Jan 2016 From: Blackpool Posts: 95 Thanks: 2 Convergence question for part one of the question it asks me to find the radius of convergence of the series $\sum_{n=0}^{\infty}\frac{x^n}{n}$ and i got the answer |x|<1 so radius of convergence R=1 for the second part of the question I am asked to show that whenever |x|
 November 14th, 2017, 09:04 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,011 Thanks: 1044 the idea here is that provided the series converges it represents the integral of $f^\prime(x) = \dfrac{1}{1-x}$ This is easily integrated to $f(x) = -\ln(1-x)$ So you have to show that $\sum \limits_{n=1}^\infty \dfrac{x^n}{n} = -\ln(1-x)$ can you do that? note: you have the series incorrectly starting at $n=0$ The 0th term is $-ln(1-0) = 0$ Thanks from v8archie and Jaket1
 November 14th, 2017, 10:24 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,331 Thanks: 2457 Math Focus: Mainly analysis and algebra I would think that you could demonstrate that $f'(x)= \frac1{1-x}$ by differentiating term by term and then demonstrating that the result is equivalent to the MacLaurin series for $\frac1{1-x}$. Thanks from Country Boy, romsek and Jaket1
 November 15th, 2017, 04:13 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,248 Thanks: 887 Or simply observing that $\frac{1}{1- x}$ is the sum of the geometric series $\displaystyle 1+ x+ x^2+ x^3+ \cdot\cdot\cdot$. as long as |x|< 1.
November 15th, 2017, 02:44 PM   #5
Member

Joined: Jan 2016
From: Blackpool

Posts: 95
Thanks: 2

Quote:
 Originally Posted by romsek the idea here is that provided the series converges it represents the integral of $f^\prime(x) = \dfrac{1}{1-x}$ This is easily integrated to $f(x) = -\ln(1-x)$ So you have to show that $\sum \limits_{n=1}^\infty \dfrac{x^n}{n} = -\ln(1-x)$ can you do that? note: you have the series incorrectly starting at $n=0$ The 0th term is $-ln(1-0) = 0$

Hi I used the maclaurin series to prove that the sum on x^n/n is approximately equal to -ln(1-x) and it works out when n>or equal to 1, thanks for the help guys.

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