Re: Sum of cos(t log(n))
- From: David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx>
- Date: Mon, 15 Jun 2009 13:20:14 -0500
On Mon, 15 Jun 2009 11:35:57 -0400, A N Niel
<anniel@xxxxxxxxxxxxxxxxxxxxx> wrote:
In article <4a366712$0$15325$426a34cc@xxxxxxxxxxxx>, Fatal
<fatal@xxxxxxxx> wrote:
José Carlos Santos a écrit :
Hi all,
Consider the series sum_n cos(t*log(n)), where _t_ is a real number. Of
course, if t = 0 this is the harmonic series, which diverges. What's the
behavior of the series for other values of _t_?
Best regards,
Jose Carlos Santos
If you meant "sum_n cos(t*log(n))/n", you may use the following result:
if f' is integrable at infinity, the series sum_n f(n) is convergent iff
the integral of f is convergent at infinity.
Interesting. Do you have a reference for this?
Assuming he means "f' is Lebesgue integrable at infinity" this is
pretty easy. In fact we only need to assume that f has bounded
variation. In that case the sum of the variation of f on [n, n+1]
is finite, hence
sum_n sup_{x in [n,n+1]} |f(n) - f(x)| < infinity
so
sum_n |f(n) - int_n^{n+1} f| < infinity.
f' is (bounded)/n^2, so integrable at infinity.
integral of f is sin(t log(n))/t, not convergent at n -> infinity, so
we conclude sum f is not convergent.
.
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- Sum of cos(t log(n))
- From: José Carlos Santos
- Re: Sum of cos(t log(n))
- From: Fatal
- Re: Sum of cos(t log(n))
- From: A N Niel
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