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I am trying to prove that $$\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k} = \sum_{k=1}^n \frac{1}{n+k}$$ for $n \in \mathbb{N}$.

My approach is to prove this by induction and this is what I got so far:

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

However, I am not aware how to proceed from here to get to: $$\sum_{k=1}^{n+1} \frac{1}{(n+1)+k}?$$

I would greatly appreciate any advice/solutions!

amWhy
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Tony Starck
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    Hint: Write $$\frac{(-1)^{k-1}}{k}=\frac{1}{k}-\frac{1+(-1)^{k}}{2} \frac{1}{k/2} $$ and note that $\frac{1+(-1)^{k}}{2}$ is non-zero only at even integers $k$ with the value $1$. So, using the indicator function notation, $$\frac{(-1)^{k-1}}{k}=\frac{1}{k}-\frac{1}{k/2}\mathbf{1}_{{k\text{ is even}}}. $$ This will help you get the desired conclusion. – Sangchul Lee Jan 28 '20 at 23:05
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    With index rewriting we get $$\sum_{k=1}^{n+1} \frac{1}{n+1+k}=\sum_{k=2}^{n+2} \frac{1}{n+k}$$ Can you take it from here? – Maximilian Janisch Jan 28 '20 at 23:07

3 Answers3

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Separate odd and even terms,

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

sentheta
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Got it, working my way backwards using Maximilian Janisch advice: $$\sum_{k=1}^{n+1} \frac{1}{n+1+k}=\sum_{k=2}^{n+2} \frac{1}{n+k} = (\sum_{k=1}^{n} \frac{1}{n+k}) + \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} = (\sum_{k=1}^{n} \frac{1}{n+k}) + \frac{1}{2n+1} - \frac{1}{2n+2} = (\sum_{k=1}^n \frac{1}{n+k}) + \frac{(-1)^{2n}}{2n+1} + \frac{(-1)^{2n+1}}{2n+2}$$

Thanks to everyone who helped!

Tony Starck
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Call the left-hand side of the expression $f(n)$ and the right-hand side $g(n)$. To show the induction step, you need to show $f(n+1) - f(n) = g(n+1) - g(n)$, because you can add this to $f(n) = g(n)$ to get $f(n+1) = g(n+1)$. Now we have $$f(n+1) - f(n) = {1 \over 2n + 1} - {1 \over 2n + 2}$$ On the other hand, $$g(n) = \sum_{k = 1}^n {1 \over n + k} = \sum_{j= n+1}^{2n} {1 \over j}$$ The last equality holds by letting $j = n + k$. As a result, we have $$g(n+1) - g(n) = \sum_{j= n+2}^{2n + 2} {1 \over j} - \sum_{j= n+1}^{2n} {1 \over j}$$ $$= {1 \over 2n + 1} + {1 \over 2n + 2} - {1 \over n+1}$$ So the statement that $f(n+1) - f(n) = g(n+1) - g(n)$ is equivalent to the statement that $${1 \over 2n + 1} - {1 \over 2n + 2} = {1 \over 2n + 1} + {1 \over 2n + 2} - {1 \over n+1}$$ This can be verified using some simple algebra.

Zarrax
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