Wolfram says sum diverges, but Mathematica gives a numerical value for infinite sum

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1
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Take this sum for example:
$$sum_n=2^inftyfrac1log(n!)$$
Wolfram says that this does not converge by the comparison test. However, when I use Mathematica's NSum function, it returns a numerical value for the summation. Who should I trust?



NSum[1/Log[n!], n, 2, [Infinity]]=6.12902





summation wolfram-alpha-queries







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  • 1




    Stirling: $log n!sim nlog n$, and $frac1nlog n$ is not summable.
    – AccidentalFourierTransform
    3 hours ago










  • @AccidentalFourierTransform by not summable, do you mean divergent?
    – John Glenn
    3 hours ago










  • yes indeed.$$
    – AccidentalFourierTransform
    3 hours ago










  • So what exactly happens when Mathematica gives me a numerical answer?
    – John Glenn
    3 hours ago










  • About Regularization?
    – Î‘λέξανδρος Ζεγγ
    3 hours ago














up vote
1
down vote

favorite












Take this sum for example:
$$sum_n=2^inftyfrac1log(n!)$$
Wolfram says that this does not converge by the comparison test. However, when I use Mathematica's NSum function, it returns a numerical value for the summation. Who should I trust?



NSum[1/Log[n!], n, 2, [Infinity]]=6.12902





summation wolfram-alpha-queries







share|improve this question

















  • 1




    Stirling: $log n!sim nlog n$, and $frac1nlog n$ is not summable.
    – AccidentalFourierTransform
    3 hours ago










  • @AccidentalFourierTransform by not summable, do you mean divergent?
    – John Glenn
    3 hours ago










  • yes indeed.$$
    – AccidentalFourierTransform
    3 hours ago










  • So what exactly happens when Mathematica gives me a numerical answer?
    – John Glenn
    3 hours ago










  • About Regularization?
    – Î‘λέξανδρος Ζεγγ
    3 hours ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Take this sum for example:
$$sum_n=2^inftyfrac1log(n!)$$
Wolfram says that this does not converge by the comparison test. However, when I use Mathematica's NSum function, it returns a numerical value for the summation. Who should I trust?



NSum[1/Log[n!], n, 2, [Infinity]]=6.12902





summation wolfram-alpha-queries







share|improve this question













Take this sum for example:
$$sum_n=2^inftyfrac1log(n!)$$
Wolfram says that this does not converge by the comparison test. However, when I use Mathematica's NSum function, it returns a numerical value for the summation. Who should I trust?



NSum[1/Log[n!], n, 2, [Infinity]]=6.12902





summation wolfram-alpha-queries




summation wolfram-alpha-queries






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 3 hours ago









John Glenn

1285




1285







  • 1




    Stirling: $log n!sim nlog n$, and $frac1nlog n$ is not summable.
    – AccidentalFourierTransform
    3 hours ago










  • @AccidentalFourierTransform by not summable, do you mean divergent?
    – John Glenn
    3 hours ago










  • yes indeed.$$
    – AccidentalFourierTransform
    3 hours ago










  • So what exactly happens when Mathematica gives me a numerical answer?
    – John Glenn
    3 hours ago










  • About Regularization?
    – Î‘λέξανδρος Ζεγγ
    3 hours ago












  • 1




    Stirling: $log n!sim nlog n$, and $frac1nlog n$ is not summable.
    – AccidentalFourierTransform
    3 hours ago










  • @AccidentalFourierTransform by not summable, do you mean divergent?
    – John Glenn
    3 hours ago










  • yes indeed.$$
    – AccidentalFourierTransform
    3 hours ago










  • So what exactly happens when Mathematica gives me a numerical answer?
    – John Glenn
    3 hours ago










  • About Regularization?
    – Î‘λέξανδρος Ζεγγ
    3 hours ago







1




1




Stirling: $log n!sim nlog n$, and $frac1nlog n$ is not summable.
– AccidentalFourierTransform
3 hours ago




Stirling: $log n!sim nlog n$, and $frac1nlog n$ is not summable.
– AccidentalFourierTransform
3 hours ago












@AccidentalFourierTransform by not summable, do you mean divergent?
– John Glenn
3 hours ago




@AccidentalFourierTransform by not summable, do you mean divergent?
– John Glenn
3 hours ago












yes indeed.$$
– AccidentalFourierTransform
3 hours ago




yes indeed.$$
– AccidentalFourierTransform
3 hours ago












So what exactly happens when Mathematica gives me a numerical answer?
– John Glenn
3 hours ago




So what exactly happens when Mathematica gives me a numerical answer?
– John Glenn
3 hours ago












About Regularization?
– Î‘λέξανδρος Ζεγγ
3 hours ago




About Regularization?
– Î‘λέξανδρος Ζεγγ
3 hours ago










1 Answer
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enter image description here
As the message shows, NSum works on limited recursions. Mathematica does not consider the "mathematical converge" when it works on NSum. Therefore Mathematica finds that this sum converges too slowly, and threw out the answer after MaxRecursion.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    enter image description here
    As the message shows, NSum works on limited recursions. Mathematica does not consider the "mathematical converge" when it works on NSum. Therefore Mathematica finds that this sum converges too slowly, and threw out the answer after MaxRecursion.






    share|improve this answer
























      up vote
      3
      down vote



      accepted










      enter image description here
      As the message shows, NSum works on limited recursions. Mathematica does not consider the "mathematical converge" when it works on NSum. Therefore Mathematica finds that this sum converges too slowly, and threw out the answer after MaxRecursion.






      share|improve this answer






















        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        enter image description here
        As the message shows, NSum works on limited recursions. Mathematica does not consider the "mathematical converge" when it works on NSum. Therefore Mathematica finds that this sum converges too slowly, and threw out the answer after MaxRecursion.






        share|improve this answer












        enter image description here
        As the message shows, NSum works on limited recursions. Mathematica does not consider the "mathematical converge" when it works on NSum. Therefore Mathematica finds that this sum converges too slowly, and threw out the answer after MaxRecursion.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 3 hours ago









        t-smart

        893114




        893114



























             

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