Alternating Harmonic Series

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Hello, over the summer I was checking out this infinite series, the alternating harmonic series, and was able to reach two contradictory answers.

Let the sum of the series be S:

$S = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + - +...$

Group the terms in to the odd and even fractions:

$S = \left(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} +...\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} +...\right)$

Now by the distributive property, pull out a half:

$S = \left(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} +...\right) - \frac{1}{2}\left(\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} +...\right)$

Group the subtracted group now into odds and evens:

$S = \left(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} +...\right) - \frac{1}{2}\left[\left(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} +...\right) + \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} +...\right)\right]$

Now seperate those groups:

$S = \left(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} +...\right) - \frac{1}{2}\left(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} +...\right) - \frac{1}{2}\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} +...\right)$

Adding the odds together, you get:

$S = \frac{1}{2}\left(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} +...\right) - \frac{1}{2}\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} +...\right)$

Multiply both sides of the equation by 2:

$2S = \left(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} +...\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} +...\right)$

Combining this equation and the first equation, we deduce that:

$2S = S$

$S = 0$

However, there are arguements that the sum of the alternating harmonic series is $ln(2)$. If you know taylor series, you can verify that. Here is another arguement:

$f(x) = \frac{x^{1}}{1} - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \frac{x^{5}}{5} -+...$ {1}

If we can find the value for $f(1)$ then our problem will be solved.

Differentiate $f(x)$ so:

$f'(x) = 1 - x^{1} + x^{2} - x^{3} + x^{4} - x^{5} +-...$ {2}

Find $xf'(x)$:

$xf'(x) = x^{1} - x^{2} + x^{3} - x^{4} + x^{5} -+...$ {3}

Adding {2} and {3} together, all but the first 1 in {2} negate each

other:

$f'(x) + xf'(x) = (1+x)f'(x) = 1$ {4}

Dividing both sides of {4} by the $(1+x)$ term:

$f'(x) = 1/(1+x)$ {5}

Now integrate both sides of equation {5}:

$f(x) = ln(1+x) + C$ {6}

You can deduce that $C = 0$ by stating $f(0) = 0$, as known from {1}, thus:

$f(x) = ln(1+x)$ {7}

Thus $f(1) = ln(1+1) = ln(2)$

Though there are several proofs to show that the series adds up to $ln2$, why would the first proof I showed you that it equals 0 be wrong?

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Probably because you're doing things with the series that aren't necessarily allowed. Re-arranging infinite sums is tricky because the alternating series isn't absolutely convergent; i.e. [imath]\sum a_n[/imath] converges but [imath]\sum |a_n| = \sum \frac{1}{n}[/imath] doesn't converge. You can quite happily re-arrange terms in the series to get two completely different answers.

I'll post later, but I'm going to have to look up some results first

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You aren't allowed to rearrange terms in an alternative series. Don't ask me why :\

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Probably because you're doing things with the series that aren't necessarily allowed. Re-arranging infinite sums is tricky because the alternating series isn't absolutely convergent; i.e. [imath]\sum a_n[/imath] converges but [imath]\sum |a_n| = \sum \frac{1}{n}[/imath] doesn't converge. You can quite happily re-arrange terms in the series to get two completely different answers.

I'll post later' date=' but I'm going to have to look up some results first [/quote']

Okay awesome, I'm looking foward to it.

You aren't allowed to rearrange terms in an alternative series. Don't ask me why :\

Um....... I'm trying to resist why........soo...ahhhhhhhh....ummmm....errrrrr... for what reason?

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The proof is a little tricky. If I remember correctly you have to consider partial sums then do some odd re-arrangement to get the limit out. I'll post again tomorrow when I can actually find my proof of it

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Are you looking for a proof for the fact that you can get any limit by rearranging (in some cases, of course) or that you're allowed to rearrange when dealing with absolute convergent series?

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from my first year lecture notes

from a random lecture note from internet

Still doesn't explain why though :\' date=' I guess I'll have to google it up.

wow this is a ton of good info, thanks!

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