Discuss the convergence of the following series whose nth term is given below:2^n-1/3^n+1

Question

This is in the comparison test

 

Answer 1

I am assuming that you mean 2^n-1/3ⁿ⁺¹

 

Let a_n = 2^n-1/3ⁿ⁺¹, b_n = 2ⁿ/3ⁿ   a_n and b_n are positive for n≥ 1

 

a_n/b_n = (2^n-1/3ⁿ⁺¹)(3ⁿ/2ⁿ) = 1/6

 

lim_n→∞(a_n/b_n) = lim_n→∞(1/6) = 1/6 > 0

 

∑_{(from n=1 to ∞)}b_n is a convergent geometric series with r  = 2/3

 

So, since a_n, b_n >0, ∑_{(n=1 to ∞)}b_n converges, and lim_n→∞(a_n/b_n)>0, then by the Limit Comparison Test, ∑_{(n=1 to ∞)}a_n also converges.  

 

 

 

 

Answer 2

San,

The easiest way I can think of to view tihis series is by re-proportioning the numerator and denominator of each term as follows:

The denominator is 3^(n-1)*3^2 or 9*3^(n-1) because (n-1)+2 = (n+1) [remember that the powers add!]

Then each term is ((2/3)^(n-1))/9

Now, that pesky discrepancy between "n-1" and "n+1" has disappeared!

Since the limit of (2/3)^infinity approaches 0, then that limit divided by 9 also approaches ?? (you figure it out!)

What have you learned about re-grouping factors in the terms of a series? Can you extend this idea a little bit, to cases where the respective powers are, say, quadratics in n (i.e. each of the form of an^2 + bn + c, but with the a,b,c different in the terms in the denominator and numerator)? Hint: try substituting a new variable to represent an appropriate offset from the old n. P.S.: this is only going to work if the expressions in n have real roots, at your level of math!

In the more general case, you have all of algebra to play around with, to do the variable transformations required to compare the numerator and denominator in a way you can recognize, so to speak. -- S.