Re: A curious result?
- To: mathgroup at smc.vnet.net
- Subject: [mg91507] Re: A curious result?
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Sun, 24 Aug 2008 07:07:16 -0400 (EDT)
- Organization: The Open University, Milton Keynes, UK
- References: <g8o89u$l8q$1@smc.vnet.net>
amzoti wrote:
> a. Sum[1/n^1.001, {n, 1 , Infinity}] = 1000.58
>
> b. Integrate[1/n^(1.001), {n, 1, Infinity}] = 1000
Only curious when one assumes that Sum[] and Integrate[] use the same
algorithm.
In[1]:= Sum[1/n^1.001, {n, 1, Infinity}]
Out[1]= 1000.58
In[2]:= Sum[1/n^(1001/1000), {n, 1, Infinity}]
% // N
Out[2]= Zeta[1001/1000]
Out[3]= 1000.58
In[4]:= Zeta[1.001]
Out[4]= 1000.58
In[5]:= Integrate[1/n^(1.001), {n, 1, Infinity}]
Out[5]= 1000.
In[6]:= Integrate[1/n^(1001/1000), {n, 1, Infinity}]
Out[6]= 1000
In[7]:= NIntegrate[1/n^(1.001), {n, 1, Infinity}]
During evaluation of In[7]:= NIntegrate::slwcon: Numerical \
integration converging too slowly; suspect one of the following: \
singularity, value of the integration is 0, highly oscillatory \
integrand, or WorkingPrecision too small. >>
During evaluation of In[7]:= NIntegrate::ncvb: NIntegrate failed to \
converge to prescribed accuracy after 9 recursive bisections in n \
near {n} = {8.16907*10^224}. NIntegrate obtained 1000.6295219735948` \
and 5.458904209437807` for the integral and error estimates. >>
Out[7]= 1000.63
Regards,
-- Jean-Marc