Can you help me with this?

Question

The sum of the infinite series :  (1/9)+(1/18)+(1/30)+(1/45)+(1/63)+........  is ?

 

(a)  2/3

(b)  1/4

(c)  1/3

(d)  1/5

Answer 1

First you have to figure out the pattern. If you look at the denominators you will see the pattern:

 

3*3, 3*6, 3*10, 3*15, ...

 

So you can rewrite this series as:

 

1/3 * (1/3 + 1/6 + 1/10 + 1/15 + 1/21 + ...)

 

The sequence of denominators is neither arithmetic nor geometric. However, if you write out the differences between them, you will get:

 

3, 4, 5, 6, ...

 

which is an arithmetic sequence. This means that the denominators form a quadratic sequence (we say that that the "second difference" is constant). In other words, the denominator is of the form

 

an² + bn + c

 

You can figure out the values of a, b, and c by substituting in three numbers in the sequence and solving the system of equations. You will find that a = 1/2, b = 3/2, and c = 1.

 

Therefore the series is:

 

(1/3) * 1 / [(1/2)n² + (3/2)n + 1]

 

which can be rewritten

 

(2/3) * 1 / [n² + 3n + 2]

 

Factor the bottom to get

 

(2/3) * 1 / [(n+2)(n+1)]

 

When you see this you should think of a telescoping series. Find the partial fraction decomposition to get:

 

(2/3) * [ 1/(n+1) - 1/(n+2) ]

 

We are taking the sum of this from n = 1 to infinity. For telescoping series you can just write out the first few terms and then the Nth term, and take the limit. You will find that everything cancels except 1/2 in the beginning; the -1/(N+2) at the end will go to zero as n goes to infinity. Therefore the answer is

 

(2/3) * (1/2) = 1/3, or c.