November 14th, 2017, 09:31 AM  #1 
Member Joined: Jan 2016 From: Blackpool Posts: 56 Thanks: 1  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<R , f'(x)=1/(1x) for all such x. Could anyone give me a hint to the second part and also check if i have done the first part correct, thanks! 
November 14th, 2017, 10:04 AM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,602 Thanks: 816 
the idea here is that provided the series converges it represents the integral of $f^\prime(x) = \dfrac{1}{1x}$ This is easily integrated to $f(x) = \ln(1x)$ So you have to show that $\sum \limits_{n=1}^\infty \dfrac{x^n}{n} = \ln(1x)$ can you do that? note: you have the series incorrectly starting at $n=0$ The 0th term is $ln(10) = 0$ 
November 14th, 2017, 11:24 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,031 Thanks: 2342 Math Focus: Mainly analysis and algebra 
I would think that you could demonstrate that $f'(x)= \frac1{1x}$ by differentiating term by term and then demonstrating that the result is equivalent to the MacLaurin series for $\frac1{1x}$.

November 15th, 2017, 05:13 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,821 Thanks: 750 
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, 03:44 PM  #5  
Member Joined: Jan 2016 From: Blackpool Posts: 56 Thanks: 1  Quote:
Hi I used the maclaurin series to prove that the sum on x^n/n is approximately equal to ln(1x) and it works out when n>or equal to 1, thanks for the help guys.  

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