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

Question

Caleb Alexander · Accepted Answer

Derive general formula for terms, split into partial fractions, then sum and simplify to find that sum is #11/18#

There is no common ratio among the terms in this series, so it is not geometric.

A general formula for one of the series' terms can be obtained as follows:

Note the terms' reciprocal sequence in writing:

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

Note the order in which those differences occur:

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

Note the order in which those differences occur:

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

Once we've arrived at a constant sequence, we can use the first term of each of these sequences as coefficients to write a formula for a general term of the original series:

#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)#

Try expressing this as a partial fraction expansion like this next:

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

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

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

Coefficients are equated to yield:

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

Hence #A = 1/3# and #B= -1/3#

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#