Let S(n) be the sum of the first 2n terms: S(n) = 1 - 1/2 + 1/4 - 1/5 + 1/7 - 1/8 + ... + 1/(3n-2) - 1/(3n-1)
The the infinite sum is lim [n->inf] S(n).
A special identity can be used to turn each reciprocal into a definite integral:
Integ [0 to 1] x^(k-1) dx = 1/k
Substitute this for each term of S(n). Then S(n) =
Integ [0 to 1] x^0 dx - Integ [0 to 1] x^1 dx
+ Integ [0 to 1] x^3 dx - Integ [0 to 1] x^4 dx
+ Integ [0 to 1] x^6 dx - Integ [0 to 1] x^7 dx
...
+ Integ [0 to 1] x^(3n-3) dx - Integ [0 to 1] x^(3n-2) dx
The sum of integrals can be condensed into one integral.
S(n) = Integ [0 to 1] (1+x^3+x^6+..+x^(3n-3)) - (x+x^4+x^7+..+x^(3n-2)) dx
These are two geometric series. Using the formula to sum the two series:
S(n) = Integ [0 to 1] (1-x^(3n))/(1-x^3) - (x-x^(3n+1))/(1-x^3) dx
A little manipulation yields
S(n) = Integ [0 to 1] 1/(1+x+x^2) dx - Integ [0 to 1] x^(3n)/(1+x+x^2) dx
The first integral can be written as
Integ [0 to 1] 1/( (sqrt(3)/2)^2 + (x+1/2) ) dx
The antiderivative is 2/sqrt(3) * arctan( (x+1/2) / (sqrt(3)/2) )
Then the definite integral equals
2/sqrt(3) * arctan( (1+1/2) / (sqrt(3)/2) ) - 2/sqrt(3) * arctan( (0+1/2) / (sqrt(3)/2) )
= 2/sqrt(3) * arctan( sqrt(3) ) - 2/sqrt(3) * arctan( 1/sqrt(3) )
= 2/sqrt(3) * pi/3 - 2/sqrt(3) * pi/6
= pi/(3*sqrt(3))
The function x^(3n)/(1+x+x^2) is between 0 and x^(3n) for x>0. Then the second integral obeys the inequality
0 <= Integ [0 to 1] x^(3n)/(1+x+x^2) dx <= Integ [0 to 1] x^(3n) dx
Combining this inequality for the second integral with the constant derived prior yields
pi/(3*sqrt(3)) <= S(n) <= pi/(3*sqrt(3)) + 1/(3n+1)
lim [n->inf] 1/(3n+1) = 0, which then implies lim S(n) [n->inf] = pi/(3*sqrt(3)).
Comments: (
You must be logged in to post comments.)
