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I found this series in my calculus book: $$\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{5^nn}$$ The directions are in the title of this question, but I can't think of any functions whose power series looks anything like that when evaluated at a point. Hints are appreciated, because I like to work these out on my own.

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2 Answers 2

up vote 8 down vote accepted

$$\log(1-x)=-\sum_{n\geqslant1}\frac{x^n}n$$

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Thanks for the hint :D –  Matt R. Aug 26 at 8:36

From Did's solution, we have

$$\log(1+x)=-\sum_{n\geqslant1}\frac{(-x)^n}n=\sum_{n\geqslant1}(-1)^{n+1}\frac{x^n}n$$

Setting $x=1/5$, we obtain:

$$\log(6/5)=\sum_{n\geqslant1}(-1)^{n+1}\frac{1}{5^n n}$$

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Quote: "Hints are appreciated, because I like to work these out on my own." –  Did Aug 26 at 8:48
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@Did, I do not mean to steal the thunder from you. But I did see that there is sign mistake in yngabl's solution when compared to the result I got from Mathematica 7.0. So I thought that it might be better to present the right result. –  mike Aug 26 at 9:24
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Your comment is mistaken: you are stealing something from the OP (the right "to work these out on (their) own", as they took care to put it), nothing from me. Additionally, reacting to a misprint in an answer by posting one's own answer instantaneously is rather peculiar, if you ask me (comments are made for this kind of situation). –  Did Aug 26 at 9:26
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@Did, Thanks for the detailed comment. I just found out the phrase "Hints are appreciated, because I like to work these out on my own" from OP. I did not read it carefully the first time. And next time, I will use the comment part to raise questions if there are any misprints. Best regards –  mike Aug 26 at 9:37

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