What is the sum from n=1 to ∞ of (-1)^(n+1) π^(2n) / (2n+1)!?

Favorite Answer

sin(x) = sum from n=0 to ∞ of (-1)^n x^(2n+1) / (2n+1)!
So the series in question is equal to
1-sin(x)/x , evaluated for x = π
= 1 - sin(π)/π = 1 - 0/π = 1
So the answer is 1....Show more

The alternating series converges as T(n)->0 as n->infinity.
The sum is
S=(pi^2)/3!+(pi^4)/5!+(pi^6)/7!+(pi^8)/9!+R(4), where R(4)=the remainder after the first 4 terms.
=>
S<0.997795618+|pi^10/11!|
=>
S<0.997795618+0.002346081
=>
S<1.000141699
=>
S=1 as n->infinity....Show more