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### Tricky Infinite Series and Radius of Convergence

```Date: 04/21/2008 at 21:32:09
Subject: A Very Tricky Series (Radius of Convergence-Endpoint Test)

This problem has consumed and befuddled 5 professors, and the entire
Tutor Center (me included) at Sierra College here.  Any help on this
final portion of the problem would be greatly helpful (considering
this is all I've been trying to solve for the past week).  This has no
grade value or anything of the sort, its just a "What-If" situation.

I will not show the original problem because there is only the issue
of the limit.  But an infinite series is given, and a radius of
convergence is determined with endpoints (8/3) and (16/3).  The left
endpoint is proven to be convergent, but as much as we extrapolate the
right endpoint (16/3), we can only assume it's divergent, without any
solid proof.  Here's the series in question (I apologize about the
poor notation, I don't have MathType handy):

SUM 'n=1->inf'

(4/3)^n * [n!(1*4*7*10...{3n-2})] / [(2n)!]

I can't really write out all the work for this, but I can list some of
what has been attempted:

Ratio/Root Tests fail with "Indeterminate"
No way on the Integral Test
Simple Divergence Test is redundant-gets us nowhere
It's not an Alternating Test

The most appealing was the Limit Comparison Test (such as relating to
(1/n), which is divergent. We can, via Excel, show massive divergence
at 100 terms, but how can we algebraically show it?

I do not expect a detailed answer, just a shove in the right direction
if possible.

Thank you.

```

```
Date: 04/22/2008 at 20:00:47
From: Doctor Vogler
Subject: Re: A Very Tricky Series (Radius of Convergence-Endpoint Test)

Thanks for writing to Dr. Math.  The first thing that comes to mind is
to use

Stirling's Approximation
http://mathforum.org/library/drmath/view/55996.html

to approximate the two factorials.  Of course, the product with every
third integer isn't as nice, so we have to handle this separately.
Well, one way to get something almost as good as Stirling's
approximation is to use the idea of

Formula to Sum a Series of Square Roots
http://mathforum.org/library/drmath/view/65309.html

We can use this to prove that

S = SUM 'k=1->inf' log(3k-2)

has

-log 2 <= S - [(n - 1/6)*log(3n - 2)] - n + 1] <= 0,

which means that

(1*4*7*10...{3n-2})

is roughly (to within a multiple of 2)

(3n - 2)^(n - 1/6) * e^(-n + 1)

Combining this with Stirling's approximation (which says that n! is

(4/3)^n * [n!(1*4*7*10...{3n-2})] / [(2n)!]

is roughly (to within a constant multiple)

(4/3)^n * n^(n + 1/2) * e^-n * (3n - 2)^(n - 1/6) * e^(-n + 1)
* (2n)^(-2n - 1/2) * e^(2n)

= (4/3)^n * n^(n + 1/2) * (3n - 2)^(n - 1/6) * (2n)^(-2n - 1/2) * e

= (1/3)^n * n^(-n) * (3n - 2)^(n - 1/6) * e/sqrt(2)

= (1/(3n))^n * (3n - 2)^(n - 1/6) * e/sqrt(2)

= ((3n - 2)/(3n))^n * (3n - 2)^(-1/6) * e/sqrt(2)

= (1 - (2/3n))^n * (3n - 2)^(-1/6) * e/sqrt(2)

I'll leave it to you to verify that the limit as n goes to infinity of

(1 - (2/3n))^n

is e^(-2/3), and more generally that for n >= 1,

1/3 <= (1 - (2/3n))^n < e^(-2/3),

which means that your summand is always within a constant multiple of

(3n - 2)^(-1/6)

and now you can use the p-test, or the integral test, to verify that
this sequence is divergent.  In other words, you have

(4/3)^n * [n!(1*4*7*10...{3n-2})] / [(2n)!] > C*(3n - 2)^(-1/6)

for some positive constant C (such as C = 1/6), and the series on the
right is divergent.

back and show me what you have been able to do, and I will try to
offer further suggestions.

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

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