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$$\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k} = \sum_{k=n+1}^{2n} \frac{1}{k}$$

I am trying to prove this inductively, so I thought that I would expand the right side out of sigma form to get

$$\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k} = \frac{2}{2n(2n+1)} - \frac{1}{n}$$

which simplified to

$$\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k} = \frac{-2}{2n+1}$$

but apparently that isn't correct, can someone provide some insight into what I am doing wrong?

user378010
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  • The second equation, how did you expand it out of sigma form? – Siong Thye Goh Oct 13 '16 at 01:00
  • $$\sum_{k=1}^n k = \frac{(n)(n+1)}{2}$$ so I flipped that upside down as the k was in the denominator, replaced every k with 2n and subtracted the first term, $$\frac{(1)}{n}$$ to make the lower limit k = 1 – user378010 Oct 13 '16 at 01:08
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    That's what it looked like you were doing, and what I alluded to in my answer. You can't flip over a sum: $\frac1{a+b}$ doesn't equal $\frac1{a}+\frac1{b}.$ – Mitchell Spector Oct 13 '16 at 01:16

2 Answers2

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The right side doesn't equal what you think it equals. I think maybe you're confusing this with the different, and unrelated, formula $\sum_{k=1}^n k = \frac{n(n+1)}{2}.$ You can't get from that formula to a formula for $\sum_{k=1}^n \frac1{k}$ or anything similar.

See https://math.stackexchange.com/a/1961771/35021 for a detailed answer to your question. (The question in the other post is phrased as a limit, but the answer there solves your problem exactly.)

Community
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Mitchell Spector
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INDUCTION PROOF:

If one wishes to prove the identity using induction, we first establish a benchmark. Letting $n=1$, we find that $\sum_{k=1}^{2}\frac{(-1)^{k+1}}{k}=1/2=\sum_{k=2}^2\frac1k$.

Next, we assume that the identity works for some number $n$

$$\color{blue}{\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}}=\color{red}{\sum_{k=n+1}^{2n}\frac1k} \tag 1$$

Then, we test the validity of the relationship in $(1)$ for $n+1$. Proceeding we find that

$$\begin{align} \sum_{k=1}^{2n+2}\frac{(-1)^{k+1}}{k}&=\color{blue}{\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}}+\frac{1}{2n+1}-\frac{1}{2n+2}\\\\ &=\color{red}{\sum_{k=n+1}^{2n}\frac1k}+\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)\\\\ &=\sum_{k=n+2}^{2n+2}\frac1k+\frac{1}{n+1}-\left(\frac{1}{2n+1}+\frac{1}{2n+2}\right)+\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)\\\\ &=\sum_{k=n+2}^{2n+2}\frac1k \end{align}$$

as expected.

DIRECT PROOF:

Note that we can write

$$\begin{align} \sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}&=\sum_{k=1}^n\frac{1}{2k-1}-\sum_{k=1}^n\frac1{2k}\\\\ &=\sum_{k=1}^n\left(\frac{1}{2k-1}+\frac{1}{2n}\right)-2\sum_{k=1}^{n}\frac1{2k}\\\\ &=\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1{k}\\\\ &=\sum_{k=n+1}^{2n}\frac1k \end{align}$$

as was to be shown!

Mark Viola
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