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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?

POSTED BY: B B
2 Replies
Posted 4 months ago

This

sum((-1)^n/sqrt(n)*cos(ln(n)),{n,1,4000})

WolframAlphaInput

seems 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})

WolframAlphaInput

does 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 limit

but 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 BY: Bill Nelson
Posted 4 months ago

Anyway Bill, thanks for your help.

Might someone else have any additional ideas?

POSTED BY: B B
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