Question

What are all values of x for which the series shown converges? ​

Answer 1

Answer:

One convergence criteria that is useful here is that, if aₙ is the n-th term of this sequence, then we must have:

Iaₙ₊₁I < IaₙI

This means that the absolute value of the terms must decrease as n increases.

Then we must have:

$\frac{(x -2)^n}{n*3^n} > \frac{(x -2 )^{n+1}}{(n + 1)*3^{n+1}}$

We can write this as:

$\frac{(x -2)^n}{n*3^n} > \frac{(x -2 )^{n+1}}{(n + 1)*3^{n+1}} = \frac{(x -2)^n}{(n + 1)*3^n} * \frac{(x - 2)}{3}$

If we assume that n is a really big number, then:

n + 1 ≈ 1

And we can write:

$\frac{(x -2)^n}{n*3^n} > \frac{(x -2)^n}{(n)*3^n} * \frac{(x - 2)}{3}$

Then we have the inequality

$1 > (x - 2)/3$

And remember that this must be in absolute value, then we will have that:

-1 < (x - 2)/3 < 1

-3 < x - 2 < 3

-3 + 2 < x < 3 + 2

-1 < x < 5

The first option looks like this, but it uses the symbols ≤≥, so it is not the same as this, then the correct option will be the second.