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Find the sum of the series $\sum_{n=1}^\infty (\frac{1}{2^\frac{1}{n}}-\frac{1}{2^\frac{1}{n+1}})$. Expanding we get
$(\frac{1}{2}-\frac{1}{2^\frac{1}{2}})+(\frac{1}{2^\frac{1}{2}}-\frac{1}{2^\frac{1}{3}})+\cdots+(\frac{1}{2^\frac{1}{n}}-\frac{1}{2^\frac{1}{n+1}}). $ Cancelling equal and opposite terms we get $(\frac{1}{2}-\frac{1}{2^\frac{1}{n+1}})$. If we write it as $\frac{1}{2}-(\frac{1}{2})^\frac{1}{n+1}$, and use $\lim_{n\to\infty} x^\frac{1}{n}=0$, for $|x|<1$, then the answer that we get is $\frac{1}{2}$, but if we take the common factor $\frac{1}{2}$ out, then it becomes $\frac{1}{2}(1-(\frac{1}{2})^\frac{-1}{1+1/n})=\frac{1}{2}(1-2^\frac{1}{1+1/n}).$ As $n\to\infty$ it simplifies to $\frac{1}{2}(1-2)=-\frac{1}{2}$ OR as the author has done, $S_n=\frac{1}{2}-\frac{1}{2^\frac{1}{n+1}} \implies \lim_{n\to\infty} S_n=\frac{1}{2}-\frac{1}{1}=-\frac{1}{2}.$ What is my mistake?

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    ouch,thanx Gerry, I goofed up in that step2012-04-03

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A CW answer to remove this question from the Unanswered queue.


As Gerry Myerson remarks, we have for $0 < x < 1$:

$\lim_{n\to\infty} x^{1/n} = 1$

Other than that, your computation is correct.