Calculus II Class 03/24


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

This is a geometric series with common ratio r = x.  If | x | < 1, as we have seen, this sequence converges to 1 / (1-x).

We therefore say that the sequence 1 + x + x^2 + ... converges about x = 0, with radius of convergence 1.

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

This sequence is of the form 1 + r + r^2 + r^3 + ... for r = 2x.  This makes it a geometric series.

We therefore see that the series converges if | 2x | < 1, i.e., if | x | < 1/2.

The series 8 + 8 (x/4) + 8 (x/4)^2 + 8 (x/4)^3 + ... can be factored into the form 8 ( x/4 + (x/4)^2 + (x/4)^3 + ... ).

The sequence converges about x = 0, with radius of convergence 4.

The series 1 + (x-3) + (x-3)^2 + (x-3)^3 + ... is a geometric series with r = x-3.

We say that this series converges about x = 3, with radius of convergence 1.

The sequence shown below converges for | 2 x - 8 | < 1, which simplifies to | x - 4 | < 1/2.

The interval of convergence is centered at x = 4 and the radius of convergence is 1/2.

Does the sequence 1 + x + x^2 / 2 + x^3 / 3 + x^4 / 4 + ... converge, and if so can we find the interval of convergence?

We note that this series is not geometric.  The ratio a(n+1) / a(n) of a member of the sequence to its predecessor is not constant.

However the ratio a(n+1) / a(n) does approach a limit, which allows us to determine the radius of convergence.

We know that a series converges if the magnitude of its ratio is less than 1, so we conclude that

The same procedure can be applied to the sequence 1 + x + x^2 / 2^2 + x^3 / 3^3 + ...

The limiting value of a(n+1) / a(n) = x / n * ((n-1) / n)^(n-1) can be found as follows:

This limit occurs for any value of x.  Since the series converges when the magnitude of the ratio is < 1,  we conclude that the series converges for any x.

We say that the radius of convergence of this sequence is inifinite.

The sequence 1 + x + x^2 / 2! + x^3 / 3! + ... has ratio a(n+1) / a(n) = x / n! * (n-1)! , as shown below.

As before, the radius of convergence is inifinite.