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$\begingroup$

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.

  • 2
    Take the partial sums and see what's getting canceled ... – 2017-02-04
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    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 $1 - \frac{1}{n}.$ – 2017-02-04
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    what a ridicolous downvote – 2017-02-04
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    @Obriareos Your comment is effectively the answer I posted. – 2017-02-04
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    @ tired: Downvote? @ Dr. MV: that's true - I haven't see your answer before. – 2017-02-04

4 Answers 4

1

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.

17

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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    Telescoping series – 2017-02-04
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    Now that deserves a (+1) just for the creativity! – 2017-02-04
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    @Oliver821 very good! – 2017-02-04
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    @Dr.MV thank you very much! :) – 2017-02-04
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    Yes the hint should be obvious by now. Hint: Split the sums and solve each telescoping series individually. – 2017-02-04
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    @tired thanks!! – 2017-02-04
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    Good answer ! (+1) – 2017-02-04
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    And this put a big smile on my face. Nice picture by the way. – 2017-02-04
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    @Dr.MV the humorless guys are never far... what a poor downvote – 2017-02-04
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    Better image (I did not check the rights, though): http://www.invitinghome.com/telescopes-binoculars/img/telescope-023-1.jpg Your telescope does not telescope! – 2017-02-04
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    @CarstenS haha you are right...it was just one of the first pictures i saw when i entered "telescope" into google – 2017-02-04
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    @tired +1, and here is yet another safe-to-use [telescoping cylinder](https://commons.wikimedia.org/wiki/File:8-stage_single-acting_air_cylinder,_Ergo-Help.jpg). – 2017-02-05
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    Now we need a picture to represent a series. Maybe a large $\Sigma$ would suffice. – 2017-02-05
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    @Dr. MV: How about that one: http://combiboilersleeds.com/images/series/series-3.jpg – 2017-02-13
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    That should do it. Hilarious – 2017-02-13
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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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    Hehe ...too many very similar to a basic question. I'm deleting mine...+1 – 2017-02-04
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    @DonAntonio Yes indeed. – 2017-02-04
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It's a telescoping series.

First show:

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