# Sum of alternating series

1. Aug 11, 2008

### 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.

2. Aug 11, 2008

### Chrisas

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

3. Aug 11, 2008

### 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.

4. Aug 17, 2008

### dynamicsolo

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 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 [Broken])
http://mathworld.wolfram.com/AlternatingHarmonicSeries.html

This series also connects to some other interesting series.

Last edited by a moderator: May 3, 2017