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, 404,10,18,28,40
Write down the sequence of differences of that sequence:
color(blue)(6), 8, 10, 126,8,10,12
Write down the sequence of differences of that sequence:
color(blue)(2), 2, 22,2,2
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))an=140!+61!(n−1)+22!(n−1)(n−2)
=1/(4+6n-6+n^2-3n+2)=14+6n−6+n2−3n+2
=1/(n^2+3n)=1n2+3n
Next, try to express this as a partial fraction expansion as follows:
1/(n^2+3n) = A/n + B/(n+3)1n2+3n=An+Bn+3
= (A(n+3)+B(n))/(n^2+3n)=A(n+3)+B(n)n2+3n
= ((A+B)n+3A)/(n^2+3n)=(A+B)n+3An2+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
