# Entering a limit of sum product into Wolfram|Alpha?

Posted 4 months ago
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 Hello,I'm new to wolframalpha.com, not sure I am in the right place to ask this question, apologies if. I just want to know if Wolfram|Alpha is able to calculate this limit, to know if the sum product converges or not: $\lim_{m\rightarrow\infty} \sum_{n=1}^{m} n^{-0.5} \cos(\ln(n)) \sum_{n=1}^{m} (-1)^{n} n^{-0.5} \cos(\ln(n))$The LateX may not work...This is the "code" I enter, is it correct, please?lim( sum( n^{-0.5} * cos( ln(n) ) ,n,1,m ) * sum( (-1)^{n} *n^{-0.5} * cos(ln(n)) ,n,1,m),m,inf)What should I modify, please? Because it gives something else like "Interpreting as: cos", or it may be too much for the machine?
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Posted 4 months ago
 This sum((-1)^n/sqrt(n)*cos(ln(n)),{n,1,4000}) WolframAlphaInputseems to strongly hint that will contribute a factor of about -0.643 to the result.This sum(1/sqrt(n)*cos(ln(n)),{n,1,4000}) WolframAlphaInputdoes not strongly hint what that sum will be, but runs out of time for larger n.Wait to see if the graphs appear after it says it has run out of time doing the calculations.This lim 1/n as n->infinity is the notation that Wolfram|Alpha usually understands when trying to enter a limitbut I have not been able to coax Wolfram|Alpha into finding your limit.Part of the problem may be that Sum may be expecting your n to be an integer but that Lim is not expecting m to be an integer
Posted 4 months ago
 Anyway Bill, thanks for your help.Might someone else have any additional ideas?