MHB Ranges and Radius of convergence

[tmt1]

Supposing I have this expression:

$$\sum_{n = 1}^{\infty} \frac{x^n}{3^n}$$

and I need to find the values for x for which this converges and the radius of convergence.

I can use the radius test:

$$\lim_{{n}\to{\infty}} |\frac{{x}^{(n + 1)} 3^n}{{3}^{(n + 1)} x^n}|$$

and this equals

$$\lim_{{n}\to{\infty}} |\frac{x}{3}| = |\frac{x}{3}|$$

We need to compare the answer to 1, so if $ |\frac{x}{3}| > 1$ then it diverges, and if $ |\frac{x}{3}| < 1$ then it converges.

Thus, if $-3 < x < 3$, then the original sum converges and if $x > 3$ or $x < -3$ then the original sum diverges.

So it converges for $(-3, 3)$.

Also, we should consider the values of $-3$ and $3$.

$\sum_{n = 1}^{\infty} \frac{3^n}{3^n}$ equals $\sum_{n = 1}^{\infty} 1^n$ which diverges.

And $\sum_{n = 1}^{\infty} \frac{(-3)^n}{3^n}$ equals $\sum_{n = 1}^{\infty} (-1)^n$ which diverges.

Therefore, the original sum still converges for the values of $x = (-3, 3)$.

Is this correct?

Also, apparently the radius equals $3$, but I'm not sure why that is (or why it's not $-3$, for example)

 

[Prove It]

Supposing I have this expression:

$$\sum_{n = 1}^{\infty} \frac{x^n}{3^n}$$

and I need to find the values for x for which this converges and the radius of convergence.

I can use the radius test:

$$\lim_{{n}\to{\infty}} |\frac{{x}^{(n + 1)} 3^n}{{3}^{(n + 1)} x^n}|$$

and this equals

$$\lim_{{n}\to{\infty}} |\frac{x}{3}| = |\frac{x}{3}|$$

We need to compare the answer to 1, so if $ |\frac{x}{3}| > 1$ then it diverges, and if $ |\frac{x}{3}| < 1$ then it converges.

Thus, if $-3 < x < 3$, then the original sum converges and if $x > 3$ or $x < -3$ then the original sum diverges.

So it converges for $(-3, 3)$.

Also, we should consider the values of $-3$ and $3$.

$\sum_{n = 1}^{\infty} \frac{3^n}{3^n}$ equals $\sum_{n = 1}^{\infty} 1^n$ which diverges.

And $\sum_{n = 1}^{\infty} \frac{(-3)^n}{3^n}$ equals $\sum_{n = 1}^{\infty} (-1)^n$ which diverges.

Therefore, the original sum still converges for the values of $x = (-3, 3)$.

Is this correct?

Also, apparently the radius equals $3$, but I'm not sure why that is (or why it's not $-3$, for example)

What you have used is called the "ratio test", not the "radius test". But what you have done is completely correct, well done.

Yes, the radius of convergence is 3. We have no confirmation that x has to be a real number. If it is complex then we have convergence where $\displaystyle \begin{align*} |x| < 3 \end{align*}$, which in the complex plane is all the points that are less than 3 units away from 0 + 0i in the complex plane, in other words, a CIRCLE, which has a radius of 3 units.