How to calculate this infinite sum

There is a popular trick for this kind of sums: [math]x_n = \frac{1}{3n(3n+3)} = \frac{1}{3}\left(\frac{1}{3n}-\frac{1}{3n+3}\right)[/math]
 
[math]\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)[/math]
 
First, do a partial fraction decomposition, then we can continue from there. This should give you what @blamocur have in post#2.
[math]\frac{1}{3n(3n+3)}=\frac{A}{3n}+\frac{B}{3n+3}[/math]
 
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Consider that [imath]\dfrac{1}{3n(3n+3)}=\dfrac{1}{9n}-\dfrac{1}{9n+9}[/imath]
Then [imath]\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][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]
What does [imath]\mathop {\lim }\limits_{N \to \infty } {S_N} = ?[/imath]
 
Loki123 said:
I am not sure I understand unfortunately
Click to expand...
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:

www.khanacademy.org

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 www.khanacademy.org
 
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!
 
Subhotosh Khan said:
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:

www.khanacademy.org

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 www.khanacademy.org
Click to expand...
Watched it and managed to solve the problem. I got 1/9. Thank youu
 
Loki123 said:
Watched it and managed to solve the problem. I got 1/9. Thank youu
Click to expand...
Partial fraction decomposition:
[math]\frac{1}{3n(3n+3)}=\frac{A}{3n}+\frac{B}{3n+3}[/math]Compare the numerator:
[math]A(3n+3)+B(3n)=1[/math]For n=0,
[math]A(3)=1 \Rightarrow A=\frac{1}{3}[/math]For n= 1,
[math]\frac{1}{3}(6)+B(3)=1 \Rightarrow B=-\frac{1}{3}[/math]It follows:
[math]\sum_{k=1}^{n}\dfrac{1}{3k(3k+3)}=\sum_{k=1}^{n}\dfrac{1}{9k}-\dfrac{1}{9(k+1)}[/math]Write out the first & last few terms:
[math]k=1 \Rightarrow { \frac{1}{9} - \frac{1}{{18}}}[/math][math]k=2 \Rightarrow { \frac{1}{18} - \frac{1}{{27}}}[/math][math]\dots[/math][math]k=(n-1)\Rightarrow \frac{1}{9(n-1)}-\frac{1}{9(n)}[/math][math]k=n \Rightarrow \dfrac{1}{9n}-\dfrac{1}{9(n+1)}[/math]Notice it's a telescoping series where only the first and the last term remains.
Thus the partial sum:
[math]S_n=\sum_{k=1}^{n}\dfrac{1}{3k(3k+3)}=\frac{1}{9} - \frac{1}{9(n+1)}[/math]Furthermore, the series is convergent (by direct comparison test). Thereby, the infinite sum:
[math]\lim_{n\to \infty} S_n=\lim_{n\to \infty} [\frac{1}{9} - \frac{1}{9(n+1)}]=\frac{1}{9}[/math]
 
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