# Find the sum upto infinite terms of the series: #1/(1*3) + 2/(1*3*5) + 3/(1*3*5*7) + 4/(1*3*5*7*9).......# Using partial fractions?

##### 3 Answers

#### Explanation:

Suppose that,

It is easy to see that, the **Nr.** of the **General**

**Dr.** is, the **Product** of the first

**A.P. :**

We note that,

Now, by **Defn.,** the sum

**Enjoy Maths.!**

#### Explanation:

Observ that

In the infinite sum the second term of each item can be simplified with the first term of the following item and the second term goes to 0, so the infinite sum is the first term of the first item =

#### Explanation:

We seek:

# S = 1/(1.3) + 2/(1.3.5) + 3/(1.3.5.7) + 4/(1.3.5.7.9) + ... #

The general term is given by:

# u_n = n/(1.3.5...(2n-1)(2n+1) #

# \ \ \ \ = ((2n+1-1)/2)/(1.3.5...(2n-1)(2n+1)) #

# \ \ \ \ = 1/2 { (2n+1-1)/(1.3.5...(2n-1)(2n+1)) } #

# \ \ \ \ = 1/2 { (2n+1)/(1.3.5...(2n-1)(2n+1)) - 1/(1.3.5...(2n-1)(2n+1)) } #

# \ \ \ \ = 1/2 { 1/(1.3.5...(2n-1)) - 1/(1.3.5...(2n+1)) } #

If we write out the first few terms using this, we get:

# n=1 => u_1 = 1/2(1/1 - 1/(1.3) ) \ \ \ \ \ \ \ \ \ \ \ = 1/3#

# n=2 => u_2 = 1/2(1/(1.3) -1/(1.3.5)) \ \ \ \ \ = 2/15#

# n=3 => u_3 = 1/2(1/(1.3.5)-1/(1.3.5.7)) = 1/35#

So we can write the sum of the first n terms as follows:

# S_n = u_1 + u_2 + ... u_n #

# \ \ \ \ \ = {1/2(1/1 - 1/(1.3) )} + #

# \ \ \ \ \ \ \ \ \ \ {1/2(1/(1.3) -1/(1.3.5))} + #

# \ \ \ \ \ \ \ \ \ \ {1/2(1/(1.3.5)-1/(1.3.5.7))} + #

# \ \ \ \ \ \ \ \ \ \ vdots #

# \ \ \ \ \ \ \ \ \ \ {1/2 ( 1/(1.3.5...(2n-1)) - 1/(1.3.5...(2n+1)) ) } #

# \ \ \ \ \ = 1/2{(1/1 - 1/(1.3) ) + #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1/(1.3) -1/(1.3.5)) + #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1/(1.3.5)-1/(1.3.5.7)) + #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ vdots #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( 1/(1.3.5...(2n-1)) - 1/(1.3.5...(2n+1)) ) } #

This is a difference sum, and we can see that almost all terms will cancel with other terms, after which we are left with:

# S_n = 1/2{ 1 - 1/(1.3.5...(2n+1)) } #

And for the sum we seek we have:

# S = lim_(n rarr oo) S_n #

# \ \ = lim_(n rarr oo) 1/2{ 1 - 1/(1.3.5...(2n+1)) } #

# \ \ = 1/2{ 1 - 0 } #

# \ \ = 1/2 #

Although a surprising result we can readily confirm this via an excel spreadsheet: