Expert: Paul Klarreich - 11/16/2008

Question
Hello Mr. Klarreich, please help me with the following problem.  

Find all positive values of b for which the series Sigma b^ln(n) {n=1 to infinity} converges.

My thinking is that this is sort of like a geometric series so when b<1 but greater than 0 the series converges. However, this is not the right answer.  the correct answer from the back of the book is b<1/e.  My question is what is wrong with my reasoning and how do i get to the correct answer of b<1/e.  

Answer
Questioner:   Sinclair
Category:  Calculus
Private:  No
 
Subject:  infinite series
Question:  Hello Mr. Klarreich, please help me with the following problem.  

Find all positive values of b for which the series Sigma b^ln(n) {n=1 to infinity} converges.

My thinking is that this is sort of like a geometric series so when b<1 but greater than 0 the series converges. However, this is not the right answer.  the correct answer from the back of the book is b<1/e.  My question is what is wrong with my reasoning and how do i get to the correct answer of b<1/e.  

..........................
Hi, Sinclair,

The problem with your approach is that you are basically using a ratio test.  In that, you look at  lim a[n+1]/a[n], as  n -> inf.

In this case, the ratio --> 1, which is inconclusive in the ratio test.

However, the INTEGRAL TEST may be more promising.  Try:

{inf
| b^(ln x) dx
}1

A quick trip to THE INTEGRATOR (see below) gives this integral:

(b^Ln[x]*x)/(1 + Ln[b])

Now you want that to converge, so ignore the bottom, a constant.

The Top = b^Ln(x) * x

Write  b = e^a, for some a.

(e^a)^Ln[x] * x

e^(aLn[x] * x

e^(Ln[x])^a * x

x^a * x

= x^(a + 1)

Now as  x -> infinity, that converges if  a + 1 < 0, or a < -1, (see note 2)
or  b = e^-1 = 1/e.

(note 1) THE INTEGRATOR is at:

http://integrals.wolfram.com/index.jsp

(and it saves time for lazy old geezers like me.)

(note 2) Why do we write  a + 1 < 0, not  a + 1 <= 0?  I'll leave that to you as an exercise.

Cute problem.