How to calculate this infinite sum

[Loki123]

It's not geometric or aritmetic, what do I do?

 

[blamocur]

There is a popular trick for this kind of sums: $$x_n = \frac{1}{3n(3n+3)} = \frac{1}{3}\left(\frac{1}{3n}-\frac{1}{3n+3}\right)$$

 

[Loki123]

There is a popular trick for this kind of sums: $$x_n = \frac{1}{3n(3n+3)} = \frac{1}{3}\left(\frac{1}{3n}-\frac{1}{3n+3}\right)$$

I am not sure I understand unfortunately

 

[blamocur]

$$\frac{1}{3\cdot 6} + \frac{1}{6\cdot 9} + \frac{1}{9\cdot 12} + ... = \frac{1}{3}\left( \frac{1}{3} - \frac{1}{6} + \frac{1}{6} - \frac{1}{9} + \frac{1}{9} - \frac{1}{12} + ... \right)$$

 

[BigBeachBanana]

First, do a partial fraction decomposition, then we can continue from there. This should give you what @blamocur have in post#2.
$$\frac{1}{3n(3n+3)}=\frac{A}{3n}+\frac{B}{3n+3}$$

 

[pka]

Consider that $\dfrac{1}{3n(3n+3)}=\dfrac{1}{9n}-\dfrac{1}{9n+9}$
Then $\displaystyle{S_N} = \sum\limits_{n = 1}^N {\left[ {\frac{1}{{9n}} - \frac{1}{{9n + 9}}} \right] = \left[ {\frac{1}{9} - \frac{1}{{18}}} \right] + \left[ {\frac{1}{{18}} - \frac{1}{{27}}} \right] \ldots } \left[ {\frac{1}{{9(N - 1)}} - \frac{1}{{9(N)}}} \right] = \left[ {\frac{1}{9} - \frac{1}{{9(N)}}} \right]$$$$$$$$$
What does $\mathop {\lim }\limits_{N \to \infty } {S_N} = ?$

 

[Deleted member 4993]

I am not sure I understand unfortunately

This is called a "telescoping" series. Do a Google search. Let us know what did you find. Carefully, look at response#4 and think a bit.

reference:

Telescoping series (video) | Series | Khan Academy

Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

www.khanacademy.org

 

[Steven G]

Just to make the numbers smaller, I would have have factored out 1/9.

blamocur is correct that the method he used is the standard way to handle these problems. Add that method to your mathematical tool box!

 

[Loki123]

This is called a "telescoping" series. Do a Google search. Let us know what did you find. Carefully, look at response#4 and think a bit.

reference:

Telescoping series (video) | Series | Khan Academy

Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

www.khanacademy.org

Watched it and managed to solve the problem. I got 1/9. Thank youu

 

[BigBeachBanana]

Watched it and managed to solve the problem. I got 1/9. Thank youu

Partial fraction decomposition:
$$\frac{1}{3n(3n+3)}=\frac{A}{3n}+\frac{B}{3n+3}$$Compare the numerator:
$$A(3n+3)+B(3n)=1$$For n=0,
$$A(3)=1 \Rightarrow A=\frac{1}{3}$$For n= 1,
$$\frac{1}{3}(6)+B(3)=1 \Rightarrow B=-\frac{1}{3}$$It follows:
$$\sum_{k=1}^{n}\dfrac{1}{3k(3k+3)}=\sum_{k=1}^{n}\dfrac{1}{9k}-\dfrac{1}{9(k+1)}$$Write out the first & last few terms:
$$k=1 \Rightarrow { \frac{1}{9} - \frac{1}{{18}}}$$$$k=2 \Rightarrow { \frac{1}{18} - \frac{1}{{27}}}$$$$\dots$$$$k=(n-1)\Rightarrow \frac{1}{9(n-1)}-\frac{1}{9(n)}$$$$k=n \Rightarrow \dfrac{1}{9n}-\dfrac{1}{9(n+1)}$$Notice it's a telescoping series where only the first and the last term remains.
Thus the partial sum:
$$S_n=\sum_{k=1}^{n}\dfrac{1}{3k(3k+3)}=\frac{1}{9} - \frac{1}{9(n+1)}$$Furthermore, the series is convergent (by direct comparison test). Thereby, the infinite sum:
$$\lim_{n\to \infty} S_n=\lim_{n\to \infty} [\frac{1}{9} - \frac{1}{9(n+1)}]=\frac{1}{9}$$