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?

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

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

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?

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

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

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?

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!