What is the exact value to which the infinite series...converges?

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- Dec 15th 2009, 05:24 PMstones44Alternating Harmonic Series
What is the exact value to which the infinite series...converges?

- Dec 15th 2009, 05:25 PMBruno J.
Hint : integrate $\displaystyle \frac{1}{1-x}=1+x+x^2+x^3+...$ to get an expression for $\displaystyle \log (1-x)$ and then set $\displaystyle x=-1$.

(Note : this is an informal proof!) - Dec 15th 2009, 05:34 PMoblixps
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 PMKrizalid
of course as a conditional convergent series, we can make it to converge to any value we want.

- Dec 16th 2009, 02:16 AMHallsofIvy
- Dec 16th 2009, 07:15 AMstones44
- Dec 16th 2009, 07:21 AMBruno J.
Yes, that's what the integral is. Multiply both sides by -1, and set x=-1.