This question is related to a previous one posted by Yellow Cake:
http://answers.yahoo.com/question/index;_ylt=AkxSv...
It is easy to find a series for e (eg ∑ 1/n! ) such that the sum of the first n terms is always < e. Can a series be found for π such that the sum of the first n terms is > π? The number n can be > some number, such as, "after the first 3 terms", etc, so that thereafter, as n increases, the sum approaches π "from above", i.e. decreases as it converges towards π.
The series must contain only rational terms, otherwise we could have the trivial "series" (π + 1) - 1/2 - 1/4 - 1/8, etc. But you can start the series with any rational number, and even mix different expressions for the terms, provided that 1) any finite sum is always rational, and 2) after a certain number of terms, it montonically decreases as it converges to π.
Update:Vasek, that didn't take you too long, didn't it? But you are right, this is the easiest series to use in conjunction with Yellow Cake's problem.
Update 3:n = 1 to ∞, that is, for this version.
Answers & Comments
Verified answer
With your hint, it's pretty easy: we can start with
π/4 = arctg 1 = Σ [n=0 to +∞] (-1)^n/(2n+1).
This is a series with alternating signs (starting at a +) which is monotonically decreasing in its absolute value. Let's just take out the first term and group the subsequent ones pairwise:
Σ [n=0 to +∞] (-1)^n/(2n+1) =
= 1 + Σ [n=0 to +∞] (-1)^(n+1)/(2n+3) =
= 1 + Σ [n=0 to +∞] ( (-1)^(2n+1) / (4n+3) + (-1)^(2n+2) / (4n+5) ) =
= 1 + Σ [n=0 to +∞] (1/(4n+5) - 1/(4n+3)) =
= 1 - Σ [n=0 to +∞] 2/((4n+3)(4n+5))
As the last sum has positive terms, the whole expression is a monotonically decreasing series with a sum of π/4. π is just four times the same.
Edit: Yes, that's true. I shifted n by 1 in the beginning, if I didn't do it, I would end with a sum from 1 to +∞ in the much more elegant form you found.
Ï = [(12)^(1/2)] { â [(-3)^k]/[2k+1]} as k goes to infinity