Can you sum it?

Algebra Level 2

Sum 1 / 9 + 1 / 18 + 1 / 30 + 1 / 45 + 1 / 63 + . . . . . 1/9 + 1/18 + 1/30 + 1/45 + 1/63+ ..... upto infinity

2 3 \dfrac{2}{3} 1 3 \dfrac{1}{3} 4 9 \dfrac{4}{9}

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Ayush Garg by Ayush Garg , 22, India

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2 solutions

Curtis Clement
Apr 14, 2015

The infinite sum can be written as: 1 3 ∑ n = 2 ∞ 2 2 n ( n + 1 ) = 1 3 ∑ n = 2 ∞ ( 1 n − 1 n + 1 ) \frac{1}{3} \displaystyle\sum_{n=2}^{\infty} 2\frac{2}{n(n+1)} = \frac{1}{3} \displaystyle\sum_{n=2}^{\infty} (\frac{1}{n} - \frac{1}{n+1}) = 2 3 [ ( 1 2 − 1 3 ) + ( 1 3 − 1 4 ) ) + ( 1 4 − 1 5 ) + . . . ] = 1 3 = \frac{2}{3} [(\frac{1}{2} - \frac{1}{3}) +(\frac{1}{3} -\frac{1}{4})) +(\frac{1}{4} - \frac{1}{5}) +...] = \boxed{\frac{1}{3}}

4 Helpful 0 Interesting 1 Brilliant 0 Confused
Ayush Garg
Apr 13, 2015

The general term is 2 3 ( n + 1 ) ( n + 2 ) = 2 / 3 ∗ ( 1 n + 1 − 1 n + 2 ) \dfrac{2}{3(n+1)(n+2) } = 2/3*(\dfrac{1}{n+1} - \dfrac{1}{n+2} ) . . Now it becomes a telescoping series and take limit to infinity

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Hi!

Well , we'll all be more interested to know how you found out the general term . I think that'll make the solution "more" complete.

A Former Brilliant Member - 6 years, 2 months ago

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S = 9 + 18 + 30 + 45 + . . . . . . . . . . . T n ( 1 ) S = 9 + 18 + 30 + . . . . . . . . . . . . . T n − 1 + T n ( 2 ) ( 1 ) − ( 2 ) 0 = 9 + 9 + 12 + 15 + 18 + . . . . . . . . . . . . ( T n − T n − 1 ) − T n \displaystyle{{S=9+18+30+45+...........{ T }_{ n }\quad \quad \quad \quad \quad \quad (1)\\ S}=\quad \quad 9+18+30+.............{ T }_{ n-1 }+{ T }_{ n }\quad \quad (2)\\ (1)-(2)\\ 0=9+9+12+15+18+............({ T }_{ n }-{ T }_{ n-1 })-{ T }_{ n }}

Now using the formula for the sum of terms in AP we get,

D r = 2 3 n ( n + 1 ) {D}_{r} =\frac { 2 }{ 3n(n+1) } .

Good to see you back.!

Abhishek Sharma - 6 years, 2 months ago

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Hi!

Actually I just wanted Ayush to add it to his solution to make it more complete .

Btw sorry to disappoint you. I had come on B'ant today to solve some questions of my friend Pranjal . I'll be available every Monday though .

A Former Brilliant Member - 6 years, 2 months ago

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@A Former Brilliant Member – Well, something is better than nothing. :P

Abhishek Sharma - 6 years, 2 months ago

Wasn't the sum 1 9 + 1 18 + . . . . . . . \frac{1}{9}+\frac{1}{18}+....... ?

Anik Mandal - 5 years, 4 months ago

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