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
Sep 6, 2017

Answer:

# 1/2.#

Explanation:

Suppose that,

#S_n=1/(1*3)+2/(1*3*5)+3/(1*3*5*7)+...+t_n.#

It is easy to see that, the Nr. of the General #n^(th)# term, i.e.,

#t_n,# is #n,# whereas, its Dr. is, the Product of the first

#(n+1)# terms of the A.P. : #1,3,5,7,......,#which is, given by,

#1*3*5*7*...*(2n+1)#.

# :. t_n=n/{1*3*5*7*...*(2n+1)}, n in NN.#

We note that, #t_n=1/2[{(2n+1)-1}/{1*3*5*7*...*(2n-1)(2n+1)}]#,

#=1/2{(2n+1)/(1*3*5*...*(2n-1)(2n+1))-1/(1*3*5*...*(2n+1))}#

#:. t_n=1/2{1/(1*3*5*...*(2n-1))-1/(1*3*5*...*(2n+1))}.#

#:. S_n=t_1+t_2+t_3+...+t_n,#

#=1/2[{1/1cancel(-1/(1*3))}+{cancel(1/(1*3))cancel(-1/(1*3*5))}#
#+{cancel(1/(1*3*5))cancel(-1/(1*3*5*7))}#
#+...+{cancel(1/(1*3*5*...*(2n-1)))-1/(1*3*5*...*(2n+1))}].#

# rArr S_n=1/2[1-1/(1*3*5*...*(2n+1))].#

Now, by Defn., the sum #S# of the given infinite series, is,

#S=lim_(n to oo) S_n,#

#=lim_(n to oo)1/2[1-1/(1*3*5*...*(2n+1))],#

#=1/2[1-0],#

# rArr S=1/2.#

Enjoy Maths.!

Sep 6, 2017

Answer:

#1/2#

Explanation:

Observ that
#1/3=1/2(1-1/3)#
#2/15=1/2(1/3-1/15)#
#3/105=1/2(1/15-1/105)#
#....#

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 = #1/2#

Sep 6, 2017

Answer:

#1/2#

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:

Steve M