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What is the general term of the series: $$-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+...$$ I think that the denominator will be $(n+1)$. But what next?

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3 Answers 3

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I would propose $(-1)^n(\frac1{2n}+\frac1{2n+1})$.

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  • $\begingroup$ frankly best answer, thank you $\endgroup$
    – kurkowski
    Feb 19, 2015 at 20:09
  • $\begingroup$ A potential gotcha is expression of a partial sum; if the last denominator of your partial sum is even, then you'll need to express the last term explicitly with this expression. But for an infinite series this isn't an issue, of course. $\endgroup$
    – John
    Feb 19, 2015 at 20:22
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$$ (-1)^{\left \lfloor \dfrac{n+1}{2}\right \rfloor}\cdot\frac{1}{n+1} $$

Where $n=1,2,\cdots $.

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There probably is a simpler way, but this works:

$$\sum_{n=1}^{\infty} \frac{\sqrt{2} \sin \left[\frac{1-2n}{4}\pi\right]}{n+1}.$$

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