Series DivergenceDate: 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. Please help! 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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