0
$\begingroup$

I am trying to find the value of $$\sum_{k=1}^{\infty}{\frac{1}{(k+2)(k+3)}}$$

I do not believe it is geometric, it cannot be divided into two fractions that both converge, but it definitely does converge, and, according to WolframAlpha, to $\frac{1}{3}$. Any way I can easily show this with no more than basic calculus?

Thank you.

  • 3
    Why do you think it cannot be divided into two fractions? – 2017-02-23
  • 0
    There are two types of series you should be able to solve (at your level at least). These are geometric series and telescoping series. Now how did your teacher deal with these fractions? Or, how do you deal with fractions in general in calculus, when the denominator is already factored? – 2017-02-23
  • 0
    @carmichael561 it would just be the difference of two divergent series. Edited my question. – 2017-02-23
  • 0
    @Lucas See my comment for your apparently "divergent" series. (You aren't supposed to split them ***that*** much.) – 2017-02-23
  • 0
    @Lucas Yes but you see, if you write the first terms, that they cancel pairwise (that is what "Simply Beautiful Art" calls telescoping). – 2017-02-23

2 Answers 2

5

Note that

$$\frac{1}{k+2}-\frac{1}{k+3}=\frac{(k+3)-(k+2)}{(k+2)(k+3)}=\frac{1}{(k+2)(k+3)}$$

so

$$\begin{align}\sum_{k=1}^n \frac{1}{(k+2)(k+3)}&=\vphantom{\cfrac1{\cfrac11}}\sum_{k=1}^n \left(\frac{1}{k+2}-\frac{1}{k+3}\right)\\&=\left(\frac{1}{3}-\color{#4488dd}{\frac{1}{4}}\right)+\left(\color{#4488dd}{\frac{1}{4}}-\color{#ff4444}{\frac{1}{5}}\right)+\bigg(\color{#ff4444}{\frac{1}{5}}+\underbrace{\frac16\bigg)+\cdots+\bigg(\frac{1}{n+2}}_{\text{cancellations}}-\frac{1}{n+3}\bigg)\\&=\frac13-\frac{1}{n+3}\end{align}$$

(as the middle terms are cancelling each other)

Now, what's the value of $\sum_{k=1}^\infty \frac{1}{(k+2)(k+3)}$?

  • 0
    I hope you don't mind the edits – 2017-02-23
  • 0
    @SimplyBeautifulArt I was leaving part of the reasoning to the OP, but I don't mind at all :) – 2017-02-23
4

By partial fractions decomposition, we have $$\frac{1}{(k + 2)(k + 3)} = \frac1{k + 2} - \frac1{k + 3}$$ Now, write down the partial sum of the series: $$S_n = \left(\frac{1}{3}-\color{#4488dd}{\frac{1}{4}}\right)+\left(\color{#4488dd}{\frac{1}{4}}-\color{#ff4444}{\frac{1}{5}}\right)+\bigg(\color{#ff4444}{\frac{1}{5}}+\underbrace{\frac16\bigg)+\cdots+\bigg(\frac{1}{n+2}}_{\text{cancellations}}-\frac{1}{n+3}\bigg)$$

As you can see, all but two terms cancel pairwise, and we are left with $$S_n = \frac13 - \frac1{n + 3}$$

What do you conclude?

  • 0
    I hope noone minds these edits. I just like to make them visually pleasing :-) – 2017-02-23
  • 0
    thank you! easy to understand – 2017-02-24