# Alternating Harmonic Series

• Dec 15th 2009, 05:24 PM
stones44
Alternating Harmonic Series
What is the exact value to which the infinite series...converges?
• Dec 15th 2009, 05:25 PM
Bruno J.
Hint : integrate $\frac{1}{1-x}=1+x+x^2+x^3+...$ to get an expression for $\log (1-x)$ and then set $x=-1$.

(Note : this is an informal proof!)
• Dec 15th 2009, 05:34 PM
oblixps
take the maclaurin series for ln(1+x) which is: x - (x^2)/2 + (x^2)/3 +...(-1)^(k+1) (x^k)/k.

now plug in x=1 and notice that the right side becomes the alternating harmonic series 1 - 1/2 + 1/3 - 1/4 +... so therefore the sum of the alternating harmonic series is ln(2).
• Dec 15th 2009, 05:44 PM
Krizalid
of course as a conditional convergent series, we can make it to converge to any value we want.
• Dec 16th 2009, 02:16 AM
HallsofIvy
Quote:

Originally Posted by Krizalid
of course as a conditional convergent series, we can make it to converge to any value we want.

By rearranging the series, yes- but this was given in a specific arrangement that converges to ln(2).
• Dec 16th 2009, 07:15 AM
stones44
Quote:

Originally Posted by Bruno J.
Hint : integrate $\frac{1}{1-x}=1+x+x^2+x^3+...$ to get an expression for $\log (1-x)$ and then set $x=-1$.

(Note : this is an informal proof!)

isn't the integral of 1/(1-x) = -ln(1-x) ? so wouldn't it be -ln2
• Dec 16th 2009, 07:21 AM
Bruno J.
Yes, that's what the integral is. Multiply both sides by -1, and set x=-1.