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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.

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    $\begingroup$ Take the partial sums and see what's getting canceled ... $\endgroup$
    – rtybase
    Feb 4, 2017 at 16:34
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    $\begingroup$ So (if you're very rusty): $\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4}... + \frac{1}{n-1} - \frac{1}{n}$ e.g. equals $1 - \frac{1}{n}.$ $\endgroup$
    – Obriareos
    Feb 4, 2017 at 16:38
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    $\begingroup$ what a ridicolous downvote $\endgroup$
    – tired
    Feb 4, 2017 at 16:41
  • $\begingroup$ @Obriareos Your comment is effectively the answer I posted. $\endgroup$
    – Mark Viola
    Feb 4, 2017 at 16:43
  • $\begingroup$ @ tired: Downvote? @ Dr. MV: that's true - I haven't see your answer before. $\endgroup$
    – Obriareos
    Feb 4, 2017 at 19:11

5 Answers 5

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Hint:

enter image description here

hope you can take it from here.

As @CarstenS pointed out the picture shown here is somehow a more correct hint

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    $\begingroup$ Now that deserves a (+1) just for the creativity! $\endgroup$
    – Mark Viola
    Feb 4, 2017 at 16:38
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    $\begingroup$ @Oliver821 very good! $\endgroup$
    – tired
    Feb 4, 2017 at 16:38
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    $\begingroup$ @Dr.MV thank you very much! :) $\endgroup$
    – tired
    Feb 4, 2017 at 16:39
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    $\begingroup$ Yes the hint should be obvious by now. Hint: Split the sums and solve each telescoping series individually. $\endgroup$ Feb 4, 2017 at 16:39
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    $\begingroup$ And this put a big smile on my face. Nice picture by the way. $\endgroup$
    – Mark Viola
    Feb 4, 2017 at 16:43
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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 $(1)$.

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  • $\begingroup$ Hehe ...too many very similar to a basic question. I'm deleting mine...+1 $\endgroup$
    – DonAntonio
    Feb 4, 2017 at 16:36
  • $\begingroup$ @DonAntonio Yes indeed. $\endgroup$
    – Mark Viola
    Feb 4, 2017 at 16:37
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It's a telescoping series.

First show:

$$\sum_{k=2}^N \frac{1}{k-1}-\frac{1}{k}=1-\frac{1}{N}$$

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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.

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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.

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