Alternating Harmonic Series

[cosine]

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:

[math]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} + - +...[/math]

Group the terms in to the odd and even fractions:

[math]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)[/math]

Now by the distributive property, pull out a half:

[math]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)[/math]

Group the subtracted group now into odds and evens:

[math]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][/math]

Now seperate those groups:

[math]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)[/math]

Adding the odds together, you get:

[math]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)[/math]

Multiply both sides of the equation by 2:

[math]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)[/math]

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

[math]2S = S[/math]

[math]S = 0[/math]

 

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

[math]f(x) = \frac{x^{1}}{1} - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \frac{x^{5}}{5} -+...[/math] {1}

If we can find the value for [math]f(1)[/math] then our problem will be solved.

 

Differentiate [math]f(x)[/math] so:

[math]f'(x) = 1 - x^{1} + x^{2} - x^{3} + x^{4} - x^{5} +-...[/math] {2}

 

Find [math]xf'(x)[/math]:

[math]xf'(x) = x^{1} - x^{2} + x^{3} - x^{4} + x^{5} -+...[/math] {3}

 

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

other:

[math]f'(x) + xf'(x) = (1+x)f'(x) = 1[/math] {4}

 

Dividing both sides of {4} by the [math](1+x)[/math] term:

[math]f'(x) = 1/(1+x)[/math] {5}

 

Now integrate both sides of equation {5}:

[math]f(x) = ln(1+x) + C[/math] {6}

 

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

[math]f(x) = ln(1+x)[/math] {7}

 

Thus [math]f(1) = ln(1+1) = ln(2)[/math]

 

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

[Dave]

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

[bloodhound]

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

[cosine]

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?

[bloodhound]

from my first year lecture notes

from a random lecture note from internet

 

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

edit: some links that might help you

http://ecademy.agnesscott.edu/~lriddle/series/rearrang.pdf

http://personal.denison.edu/~feil/classes/Rearrangement.pdf

[Dave]

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

[TD]

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?

[cosine]

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.

edit: some links that might help you

http://ecademy.agnesscott.edu/~lriddle/series/rearrang.pdf

http://personal.denison.edu/~feil/classes/Rearrangement.pdf[/quote']

 

wow this is a ton of good info, thanks!