What Is the Sum of the Alternating Harmonic Series?

[bdforbes]

Find the sum of \Sigma^{\infty}_{k=1}\frac{(-1)^{k+1}}{k}

I know the series converges because the coefficients go monotonically to zero. However it's been a few years since I was taught how to sum these series, so I'm having trouble. I thought about telescoping, but no terms seem to cancel each other. I thought about maybe grouping terms together but I'm not sure how to proceed with that. I believe the sum should be ln(2), which implies those above methods aren't adequate. I was thinking maybe I need to compare the series with some other known series? A nudge in the right direction would be great, thanks.

 

[Chrisas]

I need to compare the series with some other known series? A nudge in the right direction would be great, thanks.

That is one way of solving it. You already have one possible function to compare it against right in front of you.

 

[bdforbes]

Thanks you're right, I was being a bit stupid. I didn't actually have to find that sum anyway, just show that it converged, because then it would have to be equal to the value of the function whose Taylor series I was originally calculating.
That sentence made more sense in my mind.

 

[dynamicsolo]

Find the sum of \Sigma^{\infty}_{k=1}\frac{(-1)^{k+1}}{k}

I know the series converges because the coefficients go monotonically to zero.

I think the Alternating Series Test is pretty much what you would use here; this Test comes out of dealing with series like this. Things like the Comparison or Ratio Tests aren't much help: the Ratio Test gives the ambiguous result and the general term 1/k is already so simple that there isn't much to compare it against.

One thing you could do is a variant of the proof that the absolute harmonic series diverges. Consider the partial sums

1, 1 - (1/2), 1 - (1/2) + [(1/3) - (1/4)],
1 - (1/2) + [(1/3) - (1/4)] + [(1/5) - (1/6) + (1/7) - (1/8)], ...

versus 1, 1 + (1/2), 1 + (1/2) + (1/4), 1 + (1/2) + (1/4) + (1/8), ...

It's a little bit of work, but you can show that the next block of 2^(k-1) terms in the alternating harmonic series is smaller than the kth term in the geometric series which will converge to 2.

I thought about telescoping, but no terms seem to cancel each other.

I believe there that you're thinking of series like \sum_{k=1}^{N} \frac{1}{k} - \frac{1}{k+1}

And, yes, the sum does go to ln(2) -- see, for instance:

http://en.wikipedia.org/wiki/Harmonic_series_(mathematics )
http://mathworld.wolfram.com/AlternatingHarmonicSeries.html

This series also connects to some other interesting series.