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.

  • 1
    $\begingroup$ Start with the base case for an induction, when $k=2$. $\endgroup$ – Kevin Carlson Aug 13 '12 at 5:17
  • $\begingroup$ Keyword: telescoping series. $\endgroup$ – anon Aug 13 '12 at 5:18

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:


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.

  • $\begingroup$ @BrianM.Scott: thank you so much for your details explanation! :) $\endgroup$ – DRN Aug 13 '12 at 6:06
  • $\begingroup$ @Norlyda: You’re welcome! $\endgroup$ – Brian M. Scott Aug 13 '12 at 6:08

We may also try this: $$\sum_{n=2}^{k} \frac{1}{n}= \sum_{n=2}^{\infty}\left(\frac{1}{n}-\frac{1}{n+k-1}\right) = \sum_{n=2}^{\infty} \frac{1}{n}-\sum_{n=2}^{\infty}\frac{1}{n+k-1}$$ $$\sum_{n=2}^{\infty}\frac{1}{n+k-1}=\sum_{n=2}^{\infty} \frac{1}{n}-\sum_{n=2}^{k} \frac{1}{n}=\sum_{n=k+1}^{\infty}\frac{1}{n}. $$


  • 1
    $\begingroup$ Hmm, the harmonic series... one would usually be leery of subtracting two divergent series, though. $\endgroup$ – J. M. is a poor mathematician Aug 13 '12 at 11:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.