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Proving convergence:

$$\sum_{n=1}^\infty (-1)^{n-1}\frac1n$$

Just wanted to confirm if the reason they converge is due to the fact that for n= 1, 3, 5, ... we have a positive harmonic series and for n= 2, 4, 6, ... we have a negative harmonic series? therefore the two cancel giving a convergent sum.

Does the above proof make mathematical sense?

Is it possible for the difference of two particular (not the same) harmonic series to be divergent?

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2 Answers 2

up vote 0 down vote accepted

Here I'm speaking about the series $$\sum_{n=1}^{\infty}(-1)^n\frac1n$$ (I assume this is what you intended, because $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^n}{\frac1n}=\sum_{n=1}^{\infty}(-1)^nn$ manifestly can't converge.)

Well, yes, the fact that for $n=1,3,5,\dots$ ($n=2,4,6,\dots$) you have a positive (negative) contribution to the series is the reason why it converges, but that's just an observation and definitely not a proof.

A demonstration easily follows from Leibniz criterion for alternating series, since the sequence $(a_n=\frac1n)$ is monotone decreasing and $\displaystyle \lim_{n\rightarrow\infty}a_n=0$.

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yes, I edited the original question. Currently reading up the alternating series test. –  student101 Sep 10 '12 at 9:15

The two do not cancel at all. Perhaps you want to read about Leibniz Alternating Series

About your last question: for any $\,k,m\in\Bbb R\,$ , the series

$$\sum_{n=1}^\infty\left(\frac{1}{kn}-\frac{1}{mn}\right)=\frac{m-k}{mk}\sum_{n=1}^\infty\frac{1}{n}\,\,\,\text{diverges}$$

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