Prove $\sum_{n=2}^{k} \frac{1}{n}= \sum_{n=2}^{\infty}\left(\frac{1}{n}-\frac{1}{n+k-1}\right)$ 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.
 A: 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.
A: 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.
