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Series Divergence

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Date: 03/03/99 at 17:23:22
From: Ellen Lee
Subject: Series Divergence

I am trying to show that the series sum(k = 0 -> infinity): (k/e)^k/k!
is divergent. A clue to this problem was to first show that:

(k/3)^k < k! < (k/2)^k (as k approaches infinity)

I do not know how to prove this inequality. And also, once this
inequality is proven, I was thinking maybe to divide (k/e)^k by all
terms: (k/e)^k /(k/3)^k < (k/e)^k/k! < (k/e)^k/(k/2)^k, which reduces
down to (3/e)^k < (k/e)^k/k! < (2/e)^k right?  Then maybe I could take
the sum of (3/e)^k and (2/e)^k and solve via geometric series. I really
do not know if this is the way to approach the problem at all.

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Date: 03/05/99 at 17:59:07
From: Doctor Nick
Subject: Re: Series Divergence

This problem is a good way to explore the use of a few new tools. The
first question is how to get an upper bound on k!. One way to start is
by noting that (let log stand for the natural logarithm)

log k! = sum of log j, with j running from 1 through k.

Now what we want is an upper bound for the sum. That is, we want to
show that the sum is smaller than something else, with the "something
else" something that we can calculate easily. This will give us an
upper bound for k!.

There are lots of ways to find an upper bound for a sum. One way to do
it is to compare the sum to an integral. What we can do is represent
the sum as a certain area, and then show that this area is less than
the area under a certain curve (i.e. an integral). One way to do this
is to represent each term of the sum as a rectangle sitting on the
x-axis. For instance, the log 2 term can be represented by a rectangle
that is 1 unit wide, and log 2 units high, sitting between x = 2 and
x = 3 on the x-axis. The log 3 term will be a rectangle that is 1 unit
wide, log 3 units high, sitting between 3 and 4 on the x-axis. Do this
for all of the terms in the sum. Notice that these rectangles just fit
under the curve y = log x, and are completely covered by the area under
the curve y = log x, with x running from 1 to k + 1. This shows that
the sum of the areas of the rectangles is less than the area under the
curve. The area of the rectangles is log k!, so this tells us that log
k! is less than the integral of log x with x running from 1 to k+1.

Evaluating that integral gives us

log k! < (k+1) log (k+1) - k

This gives

k! < ((k+1)^(k+1)) e^(-k)

This means that your expression (k/e)^k/k! can be bounded below:

(k/e)^k/k! = (k^k)/(k! e^k) > (k^k)/((k+1)^(k+1))

Now you just need to show that the sum of

(k^k)/((k+1)^(k+1))

is divergent. I will let you give that a try.

Write back if you need more help, or if these explanations are not
clear enough.

- Doctor Nick, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Sequences, Series

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