Simple series but my answer looks wrong

[Benny]

I have doing a question which I cannot get the answer to. I would like some help with it if it can be done.
a) Derive a Taylor series about the origin for the function f(x) = log(1+x).
b) Find the interval of convergence for the Taylor series, justifying your answer.
c) Use the result of 'a' to deduce the Taylor series about the origin of the function \arctan h\left( x \right) = \frac{1}{2}\log \frac{{\left( {1 + x} \right)}}{{\left( {1 - x} \right)}}
d) On what interval does your series in 'c' converge?

I obtained:
a) \log \left( {1 + x} \right) = \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^{n + 1} \frac{{x^n }}{n}}

b) I = (-1,1]

c) \arctan h\left( x \right) = \frac{1}{2}\log \frac{{\left( {1 + x} \right)}}{{\left( {1 - x} \right)}} = \frac{1}{2}\left( {\log \left( {1 + x} \right) - \log \left( {1 - x} \right)} \right)

<br /> \log \left( {1 + x} \right) = \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^{n + 1} \frac{{x^n }}{n}} \Rightarrow \log \left( {1 - x} \right) = - \sum\limits_{n = 1}^\infty {\frac{{x^n }}{n}} <br />

<br /> \arctan h\left( x \right) = \frac{1}{2}\sum\limits_{n = 1}^\infty {\left[ {\left( { - 1} \right)^{n + 1} \frac{{x^n }}{n} + \frac{{x^n }}{n}} \right]} <br />

<br /> = \sum\limits_{n = 1}^\infty {\left[ {\left( {\left( { - 1} \right)^{n + 1} + 1} \right)\frac{{x^n }}{{2n}}} \right]} <br />

That doesn't look right.

d) I know(or rather, remember seeing somewhere) that if you have a product of series then the radius of convergence of the product is the smaller of the radii of convergence of the individual series. I don't think that applies here through since I've got a sum of series and not a product. I'm stuck on part 'c' mainly. Can someone have a look through my answer and tell me where I'm going wrong?

 

[benorin]

Good work, almost there

I have doing a question which I cannot get the answer to. I would like some help with it if it can be done.
a) Derive a Taylor series about the origin for the function f(x) = log(1+x).
b) Find the interval of convergence for the Taylor series, justifying your answer.
c) Use the result of 'a' to deduce the Taylor series about the origin of the function \arctan h\left( x \right) = \frac{1}{2}\log \frac{{\left( {1 + x} \right)}}{{\left( {1 - x} \right)}}
d) On what interval does your series in 'c' converge?
I obtained:
a) \log \left( {1 + x} \right) = \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^{n + 1} \frac{{x^n }}{n}}
b) I = (-1,1]
c) \arctan h\left( x \right) = \frac{1}{2}\log \frac{{\left( {1 + x} \right)}}{{\left( {1 - x} \right)}} = \frac{1}{2}\left( {\log \left( {1 + x} \right) - \log \left( {1 - x} \right)} \right)
<br /> \log \left( {1 + x} \right) = \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^{n + 1} \frac{{x^n }}{n}} \Rightarrow \log \left( {1 - x} \right) = - \sum\limits_{n = 1}^\infty {\frac{{x^n }}{n}} <br />
<br /> \arctan h\left( x \right) = \frac{1}{2}\sum\limits_{n = 1}^\infty {\left[ {\left( { - 1} \right)^{n + 1} \frac{{x^n }}{n} + \frac{{x^n }}{n}} \right]} <br />
<br /> = \sum\limits_{n = 1}^\infty {\left[ {\left( {\left( { - 1} \right)^{n + 1} + 1} \right)\frac{{x^n }}{{2n}}} \right]} <br />
That doesn't look right.
d) I know(or rather, remember seeing somewhere) that if you have a product of series then the radius of convergence of the product is the smaller of the radii of convergence of the individual series. I don't think that applies here through since I've got a sum of series and not a product. I'm stuck on part 'c' mainly. Can someone have a look through my answer and tell me where I'm going wrong?

Parts a and b are correct, without question.

As for part c: The last sum is good except that \frac{{x^n }}{{2n}}} should be \frac{{x^n }}{{n}}}. From there, note that

\left( {\left( { - 1} \right)^{n + 1} + 1} \right) = \left( 1 - (-1)^{n} \right), which is 0 for even n and 2 for odd n, hence your sum reduces to

\mbox{arctanh}(x) = \sum\limits_{n = 1}^{\infty} {\left[ {\left( {\left( { - 1} \right)^{n + 1} + 1} \right)\frac{{x^n }}{{n}}} \right]} <br /> = \frac{1}{2}\sum_{n=1,3,5,...}^{\infty} (2)\frac{x^{n}}{n} = \sum_{n=1,3,5,...}^{\infty} \frac{x^{n}}{n} = \sum_{n=1}^{\infty} \frac{x^{2n-1}}{2n-1}

as it must.

As for part d, the interval of convergence of the above series is -1 &lt; x &lt; 1, that is x\in (-1,1). I obtained this by using the ratio test and testing both end points (neither of them converge).

 

[Benny]

Thanks for the help benorin.