lim(n→∞)[Σ(k = 1 to n)[(1/(2^k))√((n – k)/(n – k + 1))]] = ?

As follows:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jun07/Crazylimit2.jpg


圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jun07/Crazylimit3.jpg


圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jun07/Crazylimit4.jpg


圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jun07/Crazylimit5.jpg

The answer is correct but there is no such "finite" m.

http://x61.xanga.com/256d6032c5630131099273/b95560900.jpg

Divisors of binomial coefficients

The prime divisors of C(n, k) can be interpreted as follows: if p is a prime number and pr is the highest power of p which divides C(n, k), then r is equal to the number of natural numbers j such that the fractional part of k/pj is bigger than the fractional part of n/pj. In particular, C(n, k) is always divisible by n/gcd(n,k).

A somewhat surprising result by David Singmaster (1974) is that any integer divides almost all binomial coefficients. More precisely, fix an integer d and let f(N) denote the number of binomial coefficients C(n, k) with n < N such that d divides C(n, k). Then


Since the number of binomial coefficients C(n, k) with n < N is N(N+1) / 2, this implies that the density of binomial coefficients divisible by d goes to 1.

The first term of the series:

k=1, 1/2 [(n-1)(n)]^0.5
k=2, 1/4 [(n-2)(n-1)]^0.5
.....

For a large n, [(n-1)(n)]^0.5 ~ n

lim(n→∞)[1/2 [(n-1)(n)]^0.5]~lim(n→∞)(n/2)→∞
The series is divergent.

if you meant 1/{2^k[(n-k)(n-k+1)]^0.5}

lim(n→∞)[1/2 [(n-1)(n)]^0.5]~lim(n→∞)(1/2n)→0

sum of all term lim(n→∞)[Σ(k = 1 to n)[(1/{2^k[[(n-k)(n-k+1)]^0.5]}
~lim(n→∞)[Σ(k = 1 to n)[(1/(2^k x (n-k+1))}
=lim(n→∞)[Σ(k = 1 to n)[(1/(2^k x (n-k+1))}
=lim(n→∞)[(1/2n+1/4(n-1)+......+1/2^n)]
=0 (since n→∞ 1/2n→0,and 1/2^n→0 )

不要玩，我現在讀大學都有咁深的數！～