Question

What is the interval of convergence for this power series: the sum from n=1 to infinity of ((x-5)^n)/((n!)5^n)?

Answer 1

\[\sum_{n=1}^{\infty}\frac{(x-5)^{n}}{ (n!)5^{n} }\]

Answer 2

The answer is \[( -\infty , \infty ) \]
I just need to know how to get there.

Answer 3

Let's try using the Ratio Test ^_^

Answer 4

\[\Large \lim_{n\rightarrow \infty}\left[\frac{(x-5)^{n+1}}{(n+1)!5^{n+1}}\cdot \frac{n!5^n}{(x-5)^n}\right]\]

Answer 5

\[\Large = \lim_{n\rightarrow \infty}\left[\frac{x-5}{5(n+1)}\right]\]
Unless I'm mistaken with the cancellations :D

Answer 6

That's what I got so far

Answer 7

Hang on, let's try putting more detail into this...
\[\Large \lim_{n\rightarrow \infty}\left[\frac{\color{blue}{(x-5)^{n}}(x-5)}{(n+1)\cdot \color{red}{n!}\cdot\color{green}{5^{n}}\cdot5}\cdot \frac{\color{red}{n!}\color{green}{5^n}}{\color{blue}{(x-5)^n}}\right]\]

So, the blues, greens, and reds cancel out, leaving
\[\Large = \lim_{n\rightarrow \infty}\left[\frac{x-5}{5(n+1)}\right]\]

Answer 8

This limit goes to zero as n goes to infinity... did you notice?
Don't forget that x is taken as a constant as far as n is concerned ^_^
\[\Large = \lim_{n\rightarrow \infty}\left[\frac{x-5}{5(n+1)}\right]=\color{red}0\]

Answer 9

Which means, by the ratio test, this converges (absolutely) regardless of the value of x :D

Answer 10

(Regardless of the value of x... because the limit does not involve x )

Answer 11

So if we do the ratio test and it is less than 1, then don't solve the inequality?

Answer 12

If we do the ratio test and it is less than one, then you know that the series converges regardless of x.

If on the other hand, the limit involves x somehow, that's when you solve an inequality.

for instance, say, if the limit was \[\frac{x+1}3\] then you'd have to solve
\[\Large \left|\frac{x+1}{3}\right|<1\] to get the radius and interval of convergence :3

Answer 13

Typically you're solving the limit as n approaches infinity, so isn't x usually pulled out of the limit?

Answer 14

Anyways, I appreciate your help!

Answer 15

Sorry, must have dozed off. and just treat x as a constant and work from there :)