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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



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