(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Show that

[tex]\forall n \in \matbb{N}:~~ \sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=n+1}^{2n} \frac{1}{k}[/tex]

3. The attempt at a solution

(1) Show that it is true for n = 1:

[tex]\sum_{k=1}^{2} \frac{(-1)^{k+1}}{k} = \frac{(-1)^2}{1} +\frac{(-1)^3}{2} = 1 - 1/2 = 1/2[/tex]

[tex]\sum_{k=n+1}^{2n} \frac{1}{k} = 1/2[/tex]

(2) Show that if it is true for n = p, it is also true for n= p+1

Assume that

[tex]\sum_{k=1}^{2p} \frac{(-1)^{k+1}}{k} = \sum_{k=p+1}^{2p} \frac{1}{k}[/tex]

Now,

[tex]\sum_{k=1}^{2p} \frac{(-1)^{k+1}}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + .... - \frac{1}{2p}[/tex]

[tex]\sum_{k=1}^{2(p+1)} \frac{(-1)^{k+1}}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + .... - \frac{1}{2p} + \frac{1}{2p+1} - \frac{1}{2p+2} = \sum_{k=1}^{2p} \frac{(-1)^{k+1}}{k} + \frac{1}{2p+1} - \frac{1}{2p+2} = \sum_{k=p+1}^{2p} \frac{1}{k} + \frac{1}{2p+1} - \frac{1}{2p+2}[/tex]

So, if it could be demonstrated that

[tex]\sum_{k=p+1}^{2p} \frac{1}{k} + \frac{1}{2p+1} - \frac{1}{2p+2} = \sum_{k=p+1}^{2(p+1)} \frac{1}{k}[/tex]

then we are done with (2)? To a first approximation, this does not seem to be equal? I probably made a mistake somewhere.

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# Homework Help: Some More Induction

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