e****t
发帖数: 17914 | 1 mean of ln (1+n*t),
n=1,2,..., n is finite (can also approximate as infinite if no finite
solution...);
t is a constant | p********e
发帖数: 16048 | 2 精确的通项公式不可能的
渐进公式你去积分就可以了
【在 e****t 的大作中提到】
: mean of ln (1+n*t),
: n=1,2,..., n is finite (can also approximate as infinite if no finite
: solution...);
: t is a constant
| H******r
发帖数: 27 | 3 积分是(n+1/t)ln(1+nt)-n，可以手推，n越大越精确
求和是ln(t^(1+n)Pochhammer(1/t,1+n))，Pochhammer(x,n)=x(x+1)(x+2)...(x+n-1)
，Mathematica算出来的，还没想明白怎么手推
【在 e****t 的大作中提到】
: mean of ln (1+n*t),
: n=1,2,..., n is finite (can also approximate as infinite if no finite
: solution...);
: t is a constant
| w****g
发帖数: 727 | 4 just saw this, have you figured it out?
I can give you a good upper bound,
like nlog n-n+O(log n) +1/2t^2 logn,
when t=1, use Stirling, then take derivative with t, sum with n and
integrate with t
depends how precise you need,
lower bound could be got by some classical result of Riemann zeta function
which I forgot the detail. | w****g
发帖数: 727 | 5 divide n if you need the mean | z*******5
发帖数: 2990 | |
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