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5包子求解个数学式子# PhotoGear - 摄影器材
e*t
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
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p*e
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

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H*r
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

avatar
w*g
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.
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w*g
5
divide n if you need the mean
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z*5
6
膜拜大神们。。。
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