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I am to expand $\ln(2+x)$ as a Maclaurin series, I've got that $\ln(2+x)=\sum\limits_{n=1}^{ \infty}(-\frac{1}{2})^{n}x^{n}$. Can someone check it?

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1 Answer 1

Your right side is a geometric progression. You know how to find the sum of one of those? And then see if it's the same as the left side?

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Well as far as I see the sum will depend on the value of x namely for $2 \leq abs(x)$ series will be divergent. –  XYZ May 13 '12 at 12:16
    
So expansion is not correct, for if we take e.g. x=3 $ln(5)$ is finite and the right hand side is not defined. Right? –  XYZ May 13 '12 at 12:24
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The Maclaurin series for a function $f(x)$ is not guaranteed to converge for all the values of $x$ for which $f(x)$ is defined, so your observation about the left side being finite and the right side undefined is not relevant. You haven't answered my question about geometric series. Believe me, if you can't recognize a geometric series when you see one, and then find its sum, you have a snowball's chance of understanding Maclaurin series for logarithmic functions. –  Gerry Myerson May 13 '12 at 12:51
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Even if you're having difficulty remembering how to evaluate geometric series, you could just check your equation at a nice, easy value, like $x = 0$ ... –  Neal May 13 '12 at 13:22
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The series you give in a comment looks right, the series in the post does not. You can find the MacLaurin series in this case directly from the definition, derivatives are easy to compute. Or else use $\int_0^x \frac{dt}{2+t}$, and expand $1/(2+t)=(1/2)(1+t/2)$ (geometric series) and integrate term by term. –  André Nicolas May 13 '12 at 13:38

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