1. Interval of convergence

I need to determine the interval of convergence of the power series

infinity
sum of series
n=1

(x+2)^n/3n5^n

I have got as far as

Here An=1/(3n5^n), for n=1,2,...

|An+1/An|=1/(3n+1)5^n+1 times 3n5^n/1

= 1/(3n+1/n)^5 ---> 1/5 as n ----> infinity.

We have R=5, by the Ratio Test. Since a=?

Can anyone help?

2. Re: Interval of convergence

Originally Posted by Arron
I need to determine the interval of convergence of the power series

infinity
sum of series
n=1

(x+2)^n/3n5^n

I have got as far as

Here An=1/(3n5^n), for n=1,2,...

|An+1/An|=1/(3n+1)5^n+1 times 3n5^n/1

= 1/(3n+1/n)^5 ---> 1/5 as n ----> infinity.

We have R=5, by the Ratio Test. Since a=?

Can anyone help?
Shouldn't $a_n=\frac{(x+2)^n}{3n*5^n}$?

3. Re: Interval of convergence

Maybe but does (x+2)^n = 1 in this instance?

4. Re: Interval of convergence

Originally Posted by Arron
Maybe but does (x+2)^n = 1 in this instance?
What would give you that idea?

5. Re: Interval of convergence

Originally Posted by Arron
I need to determine the interval of convergence of the power series
infinity
sum of series
n=1
(x+2)^n/3n5^n
Here is what you must do.
$\lim _{n \to \infty } \sqrt[n]{{\frac{{\left| {x + 2} \right|^n}}{{3n \cdot 5^n }}}} \to ? < 1$.

Be sure to test the end points.

6. Re: Interval of convergence

If you would do it with the ratio test then you get:
$\lim_{n \to +\infty} \left|\frac{(x+2)^{n+1}}{3(n+1)\cdot 5^{n+1}}\cdot \frac{3n\cdot 5^n}{(x+2)^n}\right|$
$=\lim_{n \to +\infty} \left|\frac{(x+2)^{n+1}}{(x+2)^n}\cdot \frac{3n}{3(n+1)}\cdot \frac{5^n}{5^n\cdot 5}\right|$
$=\frac{x+2}{5}\lim_{n\to +\infty} \left|\frac{3n}{3n+1}\right|$
$=\frac{3(x+2)}{5}$

Now you have two different cases ...

7. Re: Interval of convergence

Originally Posted by Siron
If you would do it with the ratio test then you get:
$\lim_{n \to +\infty} \left|\frac{(x+2)^{n+1}}{3(n+1)\cdot 5^{n+1}}\cdot \frac{3n\cdot 5^n}{(x+2)^n}\right|$
$=\lim_{n \to +\infty} \left|\frac{(x+2)^{n+1}}{(x+2)^n}\cdot \frac{3n}{3(n+1)}\cdot \frac{5^n}{5^n\cdot 5}\right|$
$=\frac{x+2}{5}\lim_{n\to +\infty} \left|\frac{3n}{3n+1}\right|$
$=\frac{3(x+2)}{5}$

Now you have two different cases ...
$\lim_{n \to +\infty} \left|\frac{(x+2)^{n+1}}{3(n+1)\cdot 5^{n+1}}\cdot \frac{3n\cdot 5^n}{(x+2)^n}\right|$
$=\lim_{n \to +\infty} \left|\frac{(x+2)^{n+1}}{(x+2)^n}\cdot \frac{3n}{3(n+1)}\cdot \frac{5^n}{5^n\cdot 5}\right|$
$=\frac{x+2}{5}\lim_{n\to +\infty} \left|\frac{3n}{3(n+1)}\right|$
$=\frac{x+2}{5}$

8. Re: Interval of convergence

ratio test to determine an interval of convergence ...

$\lim_{n \to \infty} \left|\frac{(x+2)^{n+1}}{3(n+1)\cdot 5^{n+1}}\cdot \frac{3n\cdot 5^n}{(x+2)^n}\right| < 1$

$\lim_{n \to \infty} \left|\frac{(x+2)^{n+1}}{(x+2)^n}\cdot \frac{3n}{3(n+1)}\cdot \frac{5^n}{5^n\cdot 5}\right| < 1$

$\left|\frac{x+2}{5}\right| \lim_{n\to \infty} \frac{3n}{3(n+1)} < 1$

$\left|\frac{x+2}{5}\right| < 1$

... don't forget to check the endpoints.

9. Re: Interval of convergence

Hi!

Thanks for your help everyone. Can we use the Quotient or Squeeze Rule to check for convergence?