The question is how to prove this equality $\sum_{n=2}^{k} \frac{1}{n}= \sum_{n=2}^{\infty}\left(\frac{1}{n}-\frac{1}{n+k-1}\right)$ I wasn't sure how to start proving this.
Prove $\sum_{n=2}^{k} \frac{1}{n}= \sum_{n=2}^{\infty}\left(\frac{1}{n}-\frac{1}{n+k-1}\right)$
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0Keyword: telescoping series. – 2012-08-13
2 Answers
Write out some terms of the righthand side:
$\begin{align*}&\left(\frac12-\frac1{k+1}\right)+\left(\frac13-\frac1{k+2}\right)+\ldots+\left(\frac1k-\frac1{2k-1}\right)\\ &+\left(\frac1{k+1}-\frac1{2k}\right)+\left(\frac1{k+2}-\frac1{2k+1}\right)+\ldots \end{align*}$
Notice that all of the negative terms are cancelled out by positive terms later in the series: $-\frac1{k+1}$ in the first term by $\frac1{k+1}$ in the $k$-th term, $-\frac1{k+2}$ in the second term by $\frac1{k+2}$ in the $(k+1)$-st term, and so on. The only terms that are left uncancelled are the positive parts of the terms in the top line above:
$\frac12+\frac13+\ldots+\frac1k\;.$
This is precisely the sum on the lefthand side.
Added: The sketch above is informal and ignores the issue of convergence of the infinite series on the righthand side of the identity. For $m\ge 2$ let $s_m=\sum_{n=2}^m\left(\frac1n-\frac1{n+k-1}\right)\;.$ For $m\ge k$ we can carry out the cancellation above to write
$\begin{align*}s_m&=\sum_{n=2}^k\frac1n-\sum_{n=m-k+2}^m\frac1{n+k-1}\\ &=\sum_{n=2}^k\frac1n-\sum_{n=m+1}^{m+k-1}\frac1n\;. \end{align*}$
Now $0\le\sum_{n=m+1}^{m+k-1}\frac1n\le\sum_{n=m+1}^{m+k-1}\frac1{m+1}=\frac{k-1}{m+1}\to 0\text{ as }m\to\infty\;,$ so $\lim_{m\to\infty}s_m=\sum_{n=2}^k\frac1n\;,$ exactly as we expected from the informal argument.
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0@Norlyda: You’re welcome! – 2012-08-13
We may also try this: $\sum_{n=2}^{k} \frac{1}{n}= \sum_{n=2}^{\infty}\left(\frac{1}{n}-\frac{1}{n+k-1}\right) = \sum_{n=2}^{\infty} \frac{1}{n}-\sum_{n=2}^{\infty}\frac{1}{n+k-1}$ $\sum_{n=2}^{\infty}\frac{1}{n+k-1}=\sum_{n=2}^{\infty} \frac{1}{n}-\sum_{n=2}^{k} \frac{1}{n}=\sum_{n=k+1}^{\infty}\frac{1}{n}. $
Q.E.D.
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1Hmm, the harmonic series... one would usually be leery of subtracting two divergent series, though. – 2012-08-13