Re: help with series!!
- From: Narcoleptic Insomniac <i_have_narcoleptic_insomnia@xxxxxxxxx>
- Date: Sun, 09 Jul 2006 02:19:36 EDT
On Jul 9, 2006 1:07 AM CT, Narcoleptic Insomniac wrote:
On Jul 8, 2006 11:21 PM CT, bill wrote:
If we assume the sum of all n (from n=1 to infinity)
an is convergent.
Let bn = sum of all ak's from k=n to infinity and
ak = an + an+1 + an+2 +
· · ·
I was looking at this again and I hope that you mean
b_n = sum_{k = n}^{oo} a_k
...where...
a_k = a_n + a_{n + 1} + a_{n + 2} + ...
...and not...
a_k = a_n + (a_n + 1) + (a_n + 2) + ...
.how can we that lim (as n goes to infinity) bn = 0.
I missed a few classes and now i'm trying to figure
this out myself.
Can anyone give me a few hints at least?
There is a test called the "limit test" or the "n'th
term test" that says:
If lim_{n -> oo} a_n =! 0, then the sum_{n = 1}^{oo} a_n
is divergent.
In other words, if the general term does not run to zero
as n goes to infinity, then the infinite sum does not
converge.
It is sometimes useful to consider the contraposition of
this test, that is:
If the sum_{n = 1}^{oo} a_n converges, then the limit
lim_{n -> oo} a_n = 0.
This latter part may be helpful to you in showing that
b_n -> 0 as n -> oo.
Regards,
Kyle Czarnecki
P.S. Note that the converse of the contraposition is
not always true. That is, it need not be the case that:
If the limit lim_{n -> oo} a_n = 0, then the summation
sum_{n = 1}^{oo} a_n is convergent.
A classic example of this is the harmonic series.
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