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#
