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November 14th, 2017, 09:31 AM   #1
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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/(1-x) 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!
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November 14th, 2017, 10:04 AM   #2
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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 $-ln(1-0) = 0$
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November 14th, 2017, 11:24 AM   #3
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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 $\frac1{1-x}$.
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November 15th, 2017, 05:13 AM   #4
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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.
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November 15th, 2017, 03:44 PM   #5
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Quote:
Originally Posted by romsek View Post
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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