Dave Posted December 14, 2003 Posted December 14, 2003 As you may or may not know, there's a very useful constant, invented by Euler which is particularly useful in applications of Number Theory and the like. It's defined to be the limit as n -> :inf: of: Dn = (:sum:i=1 to n 1/i) - log(n+1) It's also known by the Greek letter :lcgamma: (gamma). I thought I'd share this quite nice problem to show you what the value of: :sum: (-1)n+1/n (i.e. the infinite sum 1 - 1/2 + 1/3 - 1/4 + ...) converges to. The proof goes something like this: :sum: i = 1 to (2n-1) (-1)i+1 = 1 - 1/2 + 1/3 - 1/4 + ... + 1/(2n-1) = 1 + 1/2 + 1/3 + ... + 1/(2n-1) - 2[1/2 + 1/4 + ... + 1/(2n-2)] = D2n-1 + log(2n) - Dn-1 - log(n) = D2n-1 - Dn-1 + log(2) Now as n -> :inf:, D2n-1 -> :lcgamma:, Dn-1 -> :lcgamma:, so :sum: (-1)i+1 = log(2) Pretty nifty, eh?
fafalone Posted December 14, 2003 Posted December 14, 2003 You have no idea how much I hate convergence of infinite series and proofs of it right now. Comprehensive calc2 exam with 1/3rd of it on series in 27 hours :/
Dave Posted December 14, 2003 Author Posted December 14, 2003 Ah, I had to do 3 workbooks on it. Yay for comparison tests, ratio lemmas, and soforth... or not
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now