Calculus II Class 03/16

Does 1 + x + x^2 + x^3 + ... converge?

If | x | < 1 this is a geometric series with constant ratio x and converges to 1 / (1-x).

Does 1 + (2x) + (2x)^2 + (2x)^3 + ... converge?

If | 2 x | < 1, i.e., | x | < 1/2, this is a geometric series with constant ratio 2 x and converges to 1 / (1- 2x).

Does 8 + 8 (x/4) + 8 (x/4)^2 + 8(x/4)^3 + ... converge?

If r = x/4 then when | r | < 1 this is a geometric series 

8 * (1 + r + r^2 + r^3 + ... ) 

with constant ratio r and converges to 8 * 1 / (1-r).  

That is, if |x/4| < 1 this series converges to 8 / (1 - (x/4)), or rearranging things a bit

if | x | < 4 the series converges to 32 / ( 4 - x) . 

The interval of convergence is | x | < 4, or -4 < x < 4.  This is an interval centered at x = 0 and extending 4 units in both directions.  We say that 4 is the radius of convergence.

What about the series 1 + (x-3) + (x-3)^2 + (x-3)^3 + ... ?

This is a geometric series 1 + r + r^2 + r^3 + ... for r = x-3, so it converges when | r | < 1, i.e., when |x - 3| < 1.

|x-3| < 1 means that

x - 3 < 1 and x - 3 > -1, or

x < 4 and x > 2.

This gives us the interval (2, 4), centered at 3 and with 'radius' 1. 

1 + (x-3) + (x-3)^2 + (x-3)^3 + ... is a power series centered at x = 3 with radius of convergence 1.

What is the interval of convergence, and the radius of convergence, of the series 1 + (2x-8) + (2x-8)^2 + (2x - 8)^3 + ... ?

The form is 1 + r + r^2 + r^3 + ... with r = 2x - 8.  This geometric series converges for | r | < 1, or | 2x - 8 | < 1.

| 2 x - 8 | < 1 is equivalent to | x - 4 | < 1/2.  This interval of convergence is centered at x = 4 and has radius 1/2.

What can we do with 1 + x + x^2/2 + x^3/3 + x^4 / 4 + ... ?

This is not geometric because x^2 / 2 is not the square of x, nor is x^3 / 3 the cube of x.

We attempt a ratio test, which as we see below tells us that lim{n -> infinity} [ a(n+1) / a(n) ] = x.

We know that a series converges as long as | lim{n -> infinity) } [ a(n+1) / a(n) ] < 1.

• We conclude that this series therefore converges for | x | < 1, i.e., on an interval centered at x = 0 with radius of convergence 1.

Apply the same reasoning to the series 1 + x + x^2 / 2^2 + x^3 / 3^3 + x^4 / 4^4 + ... 

We need to investigate the limit as n -> infinity of the expression a(n+1) / a(n).

The third term is x^2 / 2^2, the fourth is x^3 / 3^3, etc..  The power of x is one less than the number of the term and the base and exponent of the denominator both match the power of x.  

The nth term is therefore a(n) = x^(n-1) / (n-1)^(n-1). 

We get a(n+1) / a(n) = [ x^n / n^n ] / [ x^(n-1) / (n-1)^(n-1) ] = x * (n-1)^(n-1) / n^n.

(n-1)^(n-1) / n^n can be written 

(n-1)^(n-1) / [ n^(n-1) * n ], 

which we rewrite as 

[ (n-1) / n ] ^(n-1) * 1/n.

etc.

Apply the same procedure to the series 1 + x + x^2 / 2! + x^3 / 3! + ... .

a(n) = x^(n-1) / (n-1) ! ; a(n+1) = x^n / n! so

a(n+1) / a(n) = [ x^n / n! ] / [ x^(n-1) / (n-1)! ] = x ( n-1)! / n! = x / n.

So lim{x -> infinity}(a(n+1) / a(n)) = lim{x->infinity}(x / n) = 0 for any x.