Re: Limit of sum

On Dec 15, 2:22 pm, Justinver <chans...@xxxxxxxxxx> wrote:
Hi,

How to find the limit of sum((k/n)^n, k=1..n-1)   for n->infinity

Thank you for advance.

After some manipulation we get,
(1/n^n)*(1+ 2^n + 3^n + ... + (n-2)^n + (n-1)^n)

The limit(n->infinity) of first sequence(1/n^n) is zero and hence the
limit of the above series is zero.

Best Regards!
sohail.
.

Relevant Pages

  • Re: select( ) function with socket
    ... in the descriptor sets, which may be zero if the timeout expires before ... Hence its possible that maybe your timeout os set to zero. ... > king regards ...
    (Linux-Kernel)
  • Re: Adobe Flex for MV GUI
    ... stupid use of zero to describe an element position. ... whether one regards an array as a set of elements or a string. ...
    (comp.databases.pick)
  • Re: Limit of sequence
    ... >> Then as n goes to infinity, both numerator and denominator of the ... >> exponential funciton go to zero. ... >Best regards, ...
    (sci.math)
  • Re: Gordon Brown starts the printing press: £75,000,000,000
    ... add a zero to realise the effect... ... web site at www.abelard.org - news comment service, logic, economics ...
    (uk.politics.misc)
  • Re: Limit of sequence
    ... >> Then as n goes to infinity, both numerator and denominator of the ... >> exponential funciton go to zero. ... >Best regards, ...
    (sci.math)