diverge...

[mina_world]

hello sir~

sum {n=2 to 00} (log n)^(-k)
determin whether the series converges
or diverge.

----------------------------------
since log n < n, 1/(log n) > 1/n.

if 0< k <=1, {1/(log n)}^k > (1/n)^k.
so,
by comparison test, diverge.

if k=0, diverge.
if k<0, diverge.

if k>1, how do you show it ??

[Jyrki Lahtonen]

mina_world wrote:
> hello sir~
>
> sum {n=2 to 00} (log n)^(-k)
> determin whether the series converges
> or diverge.

It shouldn't be too hard for you to prove that
for any k>1: (log n)^k < n, when n is large enough
(consider the limit of (log n)/{n^(1/k)}, n -> infinity).
From this you get a similar comparison test that
you already used in the range 0<k<=1.

Good luck

[Dave L. Renfro]

mina_world wrote:

For each k > 1 the terms approach 0 slower than those of the
harmonic series (n'th term is 1/n), which is easy to show by
L'Hopital's rule, so it diverges by comparison with the harmonic
series. In fact, the series {n^(-p) * (log n)^(-q)} diverges
for all (p,q) belonging to (-oo, 1) x (-oo, oo), a result that
you might have seen in a beginning calculus course since it
is easy to show using the integral test. (It is also easy to
show using the Cauchy condensation test.) Your series is the
special case where p=0 and q is any real number.

Incidentally, the more general series I gave also diverges
for (p,q) belonging to {1} x (-oo, 1], and converges for all
other choices for (p,q) (easy to show using either the integral
test or the Cauchy condensation test). When p=1, this series
is sometimes called the logarithmic p-series. The case where
p = q = 1 gives a very slowly diverging series, a series
in which it takes roughly exp[exp(M)] many terms for the
partial sums to reach M. For example, the number of terms
needed to reach 20 is approximately the number one followed
by 100 million zeros.

Dave L. Renfro

[mina_...@hanmail.net]

Jyrki Lahtonen 작성:

i can use L'Hopital's rule.
lim {n->00} (log n)/{n^(1/k)} = lim (1/n)/[{(1/k)[n^{(1/k)-1}]]
= lim (k/n)n^(1-(1/k)) = lim k/(n^k) = 0
so, there exists N such that
if n>N => (log n)/(n^(1/k)) < 1 => log n < n^(1/k)
=> (log n)^k < n
=> 1/{(log n)^k} > 1/n
so, diverge.

thank you very much.

[eugene]

[Ronald Bruck]

In article <e7r470$efk$1...@news2.kornet.net>, mina_world
<mina_...@hanmail.net> wrote:

Let e be any positive number. You already recognize that log n^e <
n^e, so that

log n < (n^e)/e.

This is an important trick.

Therefore

1/(log n)^k > e^k/(n^(e*k)).

Now just make sure that e*k < 1. This also handles the case 0 < k (the
case k <= 0 is trivial).

--
Ron Bruck