[math-fun] In[84]:= Limit[HarmonicNumber[E^n,E^(1/n)]/n,n->∞]
Out[84]= Limit[HarmonicNumber[E^n,E^(1/n)]/n,n->\[Infinity]] looks suspiciously like In[83]:= 1-1./E Out[83]= 0.6321205588285577 In[85]:= Table[HarmonicNumber[E^n,E^(1/n)]/n/.n->9.`44^k,{k,11}]//Column Out[85]= 0.682225592958289245702463034317220139455191 0.637623065384041617378550215129083728051 0.6327312065329068054767860013575408919 0.63218839948584639769227186418818269 0.632128096567272612533554169955621 0.6321213963536978066178286646303 0.63212065188688949834913479179 0.632120569168372114161664887 0.63212055997742594644041940 0.63212055895620970815410 0.632120558842741237265 --rwg (HarmonicNumber[a, s] == Zeta[s] - HurwitzZeta[s, 1 + a])
Yes. To Euler-Maclaurin first order, Out[377]= -((1 + (1 - (E^n)^(1 - E^(1/n)))/(1 - E^(1/n)))/n) == Inactive[Sum][k^-E^((1/n)), {k, 1, E^n}]/n In[378]:= Expand[Limit[%[[1]], n -> \[Infinity]]] Out[378]= 1 - 1/E Easy enough that Mathematica should have gotten it. --rwg On Tue, Jul 12, 2016 at 4:52 AM, Bill Gosper <billgosper@gmail.com> wrote:
Out[84]= Limit[HarmonicNumber[E^n,E^(1/n)]/n,n->\[Infinity]] looks suspiciously like In[83]:= 1-1./E Out[83]= 0.6321205588285577
In[85]:= Table[HarmonicNumber[E^n,E^(1/n)]/n/.n->9.`44^k,{k,11}]//Column Out[85]= 0.682225592958289245702463034317220139455191 0.637623065384041617378550215129083728051 0.6327312065329068054767860013575408919 0.63218839948584639769227186418818269 0.632128096567272612533554169955621 0.6321213963536978066178286646303 0.63212065188688949834913479179 0.632120569168372114161664887 0.63212055997742594644041940 0.63212055895620970815410 0.632120558842741237265 --rwg
(HarmonicNumber[a, s] == Zeta[s] - HurwitzZeta[s, 1 + a])
participants (1)
-
Bill Gosper