# How do you find the sum of the infinite geometric series (1/4)+(1/10)+(1/18)+(1/28)+(1/40)+...?

##### 1 Answer

Derive general formula for terms, split into partial fractions, then sum and simplify to find that sum is

#### Explanation:

This is not a geometric series. There is no common ratio between terms.

We can derive a general formula for a term of the series as follows:

Write down the sequence of reciprocals of the terms:

#color(blue)(4), 10, 18, 28, 40#

Write down the sequence of differences of that sequence:

#color(blue)(6), 8, 10, 12#

Write down the sequence of differences of that sequence:

#color(blue)(2), 2, 2#

Having reached a constant sequence, we can write a formula for a general term of the original series using the initial term of each of these sequences as coefficients:

#a_n = 1/(color(blue)(4)/(0!) + color(blue)(6)/(1!)(n-1) + color(blue)(2)/(2!)(n-1)(n-2))#

#=1/(4+6n-6+n^2-3n+2)#

#=1/(n^2+3n)#

Next, try to express this as a partial fraction expansion as follows:

#1/(n^2+3n) = A/n + B/(n+3)#

#= (A(n+3)+B(n))/(n^2+3n)#

#= ((A+B)n+3A)/(n^2+3n)#

Equating coefficients we find:

#{(A+B = 0), (3A = 1):}#

Hence

So:

#sum_(n=1)^N 1/(n^2+3n)#

#=1/3 sum_(n=1)^N (1/n - 1/(n+3))#

#=1/3 (sum_(n=1)^N 1/n - sum_(n=1)^N 1/(n+3))#

#=1/3 (sum_(n=1)^3 1/n + sum_(n=4)^N 1/n - sum_(n=1)^(N-3) 1/(n+3) - sum_(n=N-2)^N 1/(n+3))#

#=1/3 (1/1 + 1/2 + 1/3 + color(red)(cancel(color(black)(sum_(n=4)^N 1/n))) - color(red)(cancel(color(black)(sum_(n=4)^N 1/n))) - 1/(N+1) - 1/(N+2) - 1/(N+3))#

#=1/3 (1/1 + 1/2 + 1/3 - 1/(N+1)- 1/(N+2) - 1/(N+3))#

#=1/3 (11/6 - 1/(N+1) - 1/(N+2) - 1/(N+3))#

So:

#sum_(n=1)^oo 1/(n^2+3n)#

#= lim_(N->oo) sum_(n=1)^N 1/(n^2+3n)#

#= lim_(N->oo) (1/3 (11/6 - 1/(N+1) - 1/(N+2) - 1/(N+3)))#

#= 1/3 * 11/6#

#= 11/18#