$ \sum_{k=2}^{\infty} \left( \frac{1}{k-1} - \frac{1}{k} \right) = 1$ I'm reading a proof and very rusty. I'm missing something simple I suspect. 
Have a little mercy and give me a nudge in the right direction?
$2\epsilon + \sum_{k=2}^{\infty} 2\epsilon \left( \frac{1}{k-1} - \frac{1}{k} \right) = 4\epsilon$
I can use $\sum k^{-2}=\frac{\pi^2}{6}$ to show that it is less than $6\epsilon$ which also works for the proof this is from, but how do I show it is exactly $4\epsilon$ since that is how the book does it? In other words how do I show:
$ \sum_{k=2}^{\infty} \left( \frac{1}{k-1} - \frac{1}{k} \right) = 1$
I would prefer a hint. 
 A: Note that 
$$\begin{align}
\sum_{k=2}^K\left(\frac{1}{k-1}-\frac1k\right)&=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\cdots +\left(\frac{1}{K-1}-\frac{1}{K}\right)\\\\
&=1-\frac1K \tag 1
\end{align}$$
More formally, note that if $S_K=\sum_{k=2}^K\left(\frac{1}{k-1}-\frac1k\right)$, then $S_{K+1}=S_K+\frac{1}{K}-\frac{1}{K+1}$.  So, we can use induction to prove the result given by $(1)$.
A: Hint:

hope you can take it from here. 
As @CarstenS pointed out the picture shown here is somehow a more correct hint
A: It's a telescoping series. 
First show:
$$\sum_{k=2}^N \frac{1}{k-1}-\frac{1}{k}=1-\frac{1}{N}$$
A: Fix $n\geq 2$. Then $$\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=\sum_{k=2}^n \frac{1}{k-1}-\sum_{k=2}^n\frac{1}{k}=\sum_{k=1}^{n-1}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}=1+\sum_{k=2}^{n-1}\frac{1}{k}-\sum_{k=2}^{n-1}\frac{1}{k}-\frac{1}{n}=1-\frac{1}{n}.$$
Let $n\to \infty $ and you'll get your result.
A: Well it has been done before, but I still like to give another method:
$$\sum_{k=2}^{m}\frac{1}{k-1}-\frac{1}{k}=\sum_{k=1}^{m-1}\frac{1}{k}-\sum_{k=1}^{m}\frac{1}{k}+1$$$$=\left[\sum_{k=1}^{m-1}\frac{1}{k}-\ln\left(m-1\right)\right]-\left[\sum_{k=1}^{m}\frac{1}{k}-\ln\left(m\right)\right]+1+\ln\left(\frac{m-1}{m}\right)$$
Taking the limit of the expression yields:
$$\underbrace{\left[\lim_{m\to\infty}\sum_{k=1}^{m-1}\frac{1}{k}-\ln\left(m-1\right)\right]}_{\gamma }-\underbrace{\left[\lim_{m\to\infty}\sum_{k=1}^{m}\frac{1}{k}-\ln\left(m\right)\right]}_{\gamma }+1+0=\color{red}{1}$$
Where $\gamma$ is Euler–Mascheroni constant.
