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Given the alternating harmonic series: $$ \sum^{\infty}_{n=1}{\frac{(-1)^{n+1}}{n}} $$ and if we let the sequence of its partial sums be $s_n$ then how can we express the sequence of even partial sums $s_{2n}$ and odd partial sums $s_{2n+1}$?

I did the following but am not completely sure if it is valid:

$$ s_{2n}= \sum^{2n}_{i=1}{\frac{(-1)^{i+1}}{i}}$$ $$ s_{2n+1}= \sum^{2n+1}_{i=1}{\frac{(-1)^{i+1}}{i}}$$

Then to show the first is increasing and the second is decreasing, can we just take $s_{2n+2}-s_{2n}$ and show that this is $> 0$? Similarly then taking $s_{2n+3}-s_{2n+1}$ and showing that this is $< 0$?

And to go one step further, how could we show that each of these sequences of partial sums has a limit? And then show that the limits are the same?

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  • $\begingroup$ $s_{2n} = \sum_{i=1}^n \frac{1}{2i-1}-\frac{1}{2i}=\sum_{i=1}^n \frac{1}{2i (2i-1)}$ $\endgroup$ – reuns Nov 12 '17 at 1:31
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    $\begingroup$ en.wikipedia.org/wiki/Alternating_series_test $\endgroup$ – parsiad Nov 12 '17 at 1:33
  • $\begingroup$ I don't quite get where the formula you've written comes from? $\endgroup$ – Analysis is fun Nov 12 '17 at 1:42
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What you have done is fine. Showing the increasing/decreasing nature works well. You can invoke the alternating series theorem to show the whole series has a limit because the terms are alternating in sign and decreasing monotonically to zero. If the whole series has a limit, every subseries converges to the same limit and you are done.

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  • $\begingroup$ Ah okay I see, is there a way to directly prove that $s_{2n}$ and $s_{2n+1}$ have limits? $\endgroup$ – Analysis is fun Nov 12 '17 at 1:42
  • $\begingroup$ The subseries argument works fine. Also they are monotonic and bounded above/below so have limits. $\endgroup$ – Ross Millikan Nov 12 '17 at 1:46
  • $\begingroup$ Ah yes okay I think I've got it now, thanks so much for your help! $\endgroup$ – Analysis is fun Nov 12 '17 at 1:47
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Yes, you are correct on how to show that the first is decreasing and the second is increasing.

Now, since you know they are monotone, if you can show they are bounded, you will demonstrate that they have limits. For instance you can show that $$ 1>s_{2n+1} > 1/2$$ for all $n$.

That the limits are the same follows from the fact that the difference $$ |s_{2n+1}-s_{2n}| = \frac{1}{2n+1}\to 0. $$

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  • $\begingroup$ Thanks, so we can show boundedness and then use the monotone convergence theorem? $\endgroup$ – Analysis is fun Nov 12 '17 at 1:47
  • $\begingroup$ Yeah. As others have said the alternating series test proves it all in one fell swoop but it seemed like the point of the exercise was to prove a special case of the alternating series test. $\endgroup$ – spaceisdarkgreen Nov 12 '17 at 1:52
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Since $s_{2n}= \sum^{2n}_{i=1}{\frac{(-1)^{i+1}}{i}} $,

$\begin{array}\\ s_{2n+2} &= \sum^{2n+2}_{i=1}{\frac{(-1)^{i+1}}{i}}\\ &= \sum^{2n}_{i=1}{\frac{(-1)^{i+1}}{i}}+{\frac{(-1)^{2n+1}}{2n+1}}+{\frac{(-1)^{2n+2}}{2n+2}}\\ &= s_{2n}-{\frac{1}{2n+1}}+{\frac{1}{2n+2}}\\ &= s_{2n}-{\frac{1}{(2n+1)(2n+2)}}\\ &\lt s_{2n}\\ \end{array} $

so $s_{2n}$ is decreasing.

Similarly,

$\begin{array}\\ s_{2n+3} &= \sum^{2n+3}_{i=1}{\frac{(-1)^{i+1}}{i}}\\ &= \sum^{2n+1}_{i=1}{\frac{(-1)^{i+1}}{i}}+{\frac{(-1)^{2n+2}}{2n+2}}+{\frac{(-1)^{2n+3}}{2n+3}}\\ &= s_{2n+1}+{\frac{1}{2n+2}}-{\frac{1}{2n+3}}\\ &= s_{2n+1}+{\frac{1}{(2n+2)(2n+3)}}\\ &\gt s_{2n+1}\\ \end{array} $

so $s_{2n+1}$ is increasing.

Since $s_{2n+1} =s_{2n}-\frac1{2n+1} \lt s_{2n} $, all the $s_{2n+1}$ are less than all the $s_{2n}$.

Since $|s_{2n+1}-s_{2n}| =\frac1{2n+1} \to 0 $, the sequences $(s_{2n})$ and $(s_{2n+1})$ must approach a common limit.

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