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I want to prove the following statement by induction for $n\leq 1$: \begin{align*} \sum_{k=n}^{2n-1}\frac{1}{k}= \sum_{k=1}^{2n-1}\frac{(-1)^{k+1}}{k} \end{align*}

For $n=1$ its easy to see: \begin{align*} \sum_{k=1}^{2-1}\frac{1}{k}= \sum_{k=1}^{1}\frac{1}{k}= 1 = \sum_{k=1}^{1}\frac{(-1)^{2}}{1} \end{align*} For $n\mapsto n+1$ \begin{align*} \sum_{k=n+1}^{2(n+1)-1}\frac{1}{k}= \sum_{k=n+1}^{2n+1}\frac{1}{k} = \sum_{k=n-1}^{2n-1}\frac{1} {k}+\frac{1}{n+1}+\frac{1}{n}\dots \ ? \end{align*} I am not really sure how to deal with the $k=n+1$ in the sum, to use the case that the statement holds for n. Could anybody help me? Thank you very much!

putti.123
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1 Answers1

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Call the left sum $L(n)$ and the right sum $R(n)$. Then we have

$$L(n + 1) - L(n) = \frac{1}{2n}+\frac{1}{2n+1} - \frac{1}{n},$$

and

\begin{align} R(n + 1) - R(n) &= \frac{(-1)^{2n}}{2n+1} + \frac{(-1)^{2n+2}}{2n+1} \\ &= - \frac{1}{2n} + \frac{1}{2n+1}. \end{align}

So by comparison, we need to check that

$$\frac{1}{2n} - \frac{1}{n} = - \frac{1}{2n}, $$

which is easy enough.

Therefore $L(n+1) - L(n) = R(n + 1) - R(n)$. So if $L(n) = R(n)$ then $L(n + 1) = R(n + 1)$.

Trevor Gunn
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