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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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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.2012-05-13
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    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?2012-05-13
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    I've got this $ln(2+x)=ln(2)+\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n2^{n}}x^{n}$.2012-05-13
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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.2012-05-13
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    I can recognize a geometric series and fing it's sum. So your advice is to pick some value of x and check if my series gives good enough approximation of it?2012-05-13
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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 $x = 0$ ...2012-05-13
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    As for that the series evidently works, the problem is how to check it with whole generality?2012-05-13
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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.2012-05-13
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    The correct series will give the *exact* value of the function, not an approximation, provided you evaluate at some $x$ for which the series converges. But that's not a good way to check that you have the right series. One way to check that the answer you have given in the comments is right is to differentiate both sides. If that gives you an equality, and if in addition your answer works for one value of $x$ (Neal's suggestion is good here), then your answer is correct.2012-05-13