I have to calculate the series $$\sum_{k=2}^\infty \left(\frac 1k-\frac 1{k+2}\right)$$

Using the definition: $$L = \lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=0}^na_k$$ Obviously $\lim_{n\to\infty} (\frac 1n-\frac 1{n+2})=0$, but I don't think that this is the right way to calculate the value of the series.

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    $\begingroup$ Have you ever heard the term "Telescoping" before used in a mathematical context? $\endgroup$ – JMoravitz Jun 4 at 17:42
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    $\begingroup$ Welcome to Math Stack Exchange. Before taking the limit, first try to evaluate $S_n$ using telescoping $\endgroup$ – J. W. Tanner Jun 4 at 17:48
  • $\begingroup$ I completely forgot about the telescopic series, I was too focused on using the definition $\endgroup$ – Diego Di Marzo Jun 4 at 18:09

As suggested from the comment by JMoravitz, you should write down the series explicitly and see if you can see something interesting:

$$ \sum_{2}^\infty \big( \frac{1}{k} - \frac{1}{k+2} \big) = \bigg(\frac{1}{2} - \frac{1}{4} \bigg) + \bigg(\frac{1}{3} - \frac{1}{5} \bigg) + \bigg(\frac{1}{4} - \frac{1}{6} \bigg) + \bigg(\frac{1}{5} - \frac{1}{7} \bigg) + \bigg(\frac{1}{6} - \frac{1}{8} \bigg) + \cdots + \bigg(\frac{1}{n} - \frac{1}{n+2} \bigg) + \bigg(\frac{1}{n+1} - \frac{1}{n+3} \bigg) + \bigg(\frac{1}{n+2} - \frac{1}{n+4} \bigg) + \bigg(\frac{1}{n+3} - \frac{1}{n+5} \bigg) + \cdots $$

From here, you can see which terms are cancelling with one another and which one is left.... Essentially you should see that all the terms will cancel with each other except the $\frac{1}{2}$ and $\frac{1}{3}$. Therefore, the answer is $\frac{5}{6}$.

$\textbf{Another, Different way}$ that uses limit like you wanted to is to find the partial sum! You first compute, $S_2 = \frac{1}{2} - \frac{1}{4} $, $S_3 = S_2 + \bigg( \frac{1}{3} - \frac{1}{5} \bigg)$ and so on... What you will find (and you can prove this fact by induction) is that $$ S_n = \dfrac{5n^2 + 3n -8}{6(n+1)(n+2)}$$ Now, $$ \lim_{n \to \infty} \dfrac{5n^2 + 3n -8}{6n^2 + 18n + 12} = \frac{5}{6} $$

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    $\begingroup$ Thank you so much, I was so focused on looking the limit that I forgot to write down the series. $\endgroup$ – Diego Di Marzo Jun 4 at 18:06

Let us use $1/k=\int_{0}^1 t^{k-1} dt$. Then $$S=\sum_{k=2}^{\infty} \left (\frac{1}{k}-\frac{1}{k+2} \right)= \int_{0}^{1} \sum_{k=2}^{\infty} ~[t^{k-1}- t^{k+2}]~dt =\int_{0}^{1} \frac{t-t^3}{1-t} dt=\int_{0}^{1} (t+t^2) dt=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}.$$



Try writing out the first few partial sums and look to see if anything can cancel.

Even more explicitly, looking ahead to somewhere in the middle, you will have if you expand out the summation the following:

$$\dots +\left(\dfrac{1}{50}-\color{blue}{\dfrac{1}{52}}\right)+\left(\dfrac{1}{51}-\color{red}{\dfrac{1}{53}}\right)+\left(\color{blue}{\dfrac{1}{52}}-\dfrac{1}{54}\right)+\left(\color{red}{\dfrac{1}{53}}-\dfrac{1}{55}\right)+\dots$$

Now, notice the colors I used and think about why I put colors there. Where else in the series does something like this happen? What does this imply in the end will be left over if we take a partial sum? What does this imply in the end will be left over if we consider the limit of partial sums?

The more general name for this property is "Telescoping." We refer to this series as a "Telescoping Series."


It is $\sum_{k=2}^\infty \frac1k-\frac1{k+2}=\lim_{n\to\infty}\sum_{k=2}^n \frac1k-\frac{1}{k+2}$

As being said in the comments, this series is telescoping.

If you write down some terms for $n=5$, we would get:

$\sum_{k=2}^5 \frac1k-\frac{1}{k+2}=(\frac12-\frac14)+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+(\frac{1}{5}-\frac{1}{7})$

Here we see, what is called "the series is telescoping". There is a pattern in addition and subtraction of specific summands. First we have $-\frac14$. Later in the series, we have $+\frac14$. The same goes for $\frac15$. These summands cancel each other out.

So the long summation can be 'pushed together', like a telescop. If we would look at higher values of $n$, we would see this to more effect.

That being said, we have to formalize that.

It is $\lim_{n\to\infty}\sum_{k=2}^n\frac1k-\frac{1}{k+2}=\lim_{n\to\infty}\sum_{k=2}^n\frac1k-\sum_{k=2}^n\frac{1}{k+2}$

Note here, that we have to make this step for partial sums, and can not calculate like this for infinite series, because both sums would not converge.

Now we make an index shift:

We get $\sum_{k=2}^n\frac{1}{k+2}\to\sum_{k=4}^{n+2}\frac{1}{k}$.

We want to do that, so we can subtract easily. Think about this step.

Finally we arrive at:

$\lim_{n\to\infty} \sum_{k=2}^n \frac{1}{k}-\sum_{k=4}^{n+2}\frac1k$

Now we can see, what we observed above. Namely the partial equal terms in both sums, getting added and subtracted.

To be more formal:

$\sum_{k=2}^n \frac{1}{k}-\sum_{k=4}^{n+2}\frac1k=\left(\frac12+\frac13+\sum_{k=4}^n\frac1k\right)-\left(\sum_{k=4}^n\frac{1}{k}+\frac1{n+1}+\frac{1}{n+2}\right)=\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}$

So we end up with:

$\lim_{n\to\infty}\sum_{k=2}^n \frac1k-\frac{1}{k+2}=\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}\to_{n\to\infty} \frac12+\frac13-0-0=\frac{5}{6}$


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