Ratio Test

[Jason76]

\(\displaystyle \lim n \rightarrow \infty[\dfrac{a_{n + 1}}{a_{n}}] = L\)

Applying this formula to

Summation sign with \(\displaystyle \infty\) on top and \(\displaystyle n = 1\) on the bottom \(\displaystyle \dfrac{4^{n}}{n!}\)

\(\displaystyle \lim n \rightarrow \infty[\dfrac{\dfrac{4^{n+1}}{(n + 1)^{1}}}{\dfrac{4^{n}}{n!}}]\) Any hints on what is going on here?

 

[pka]

\(\displaystyle \lim n \rightarrow \infty[\dfrac{a_{n + 1}}{a_{n}}] = L\)

Applying this formula to

Summation sign with \(\displaystyle \infty\) on top and \(\displaystyle n = 1\) on the bottom \(\displaystyle \dfrac{4^{n}}{n!}\)

\(\displaystyle \lim n \rightarrow \infty[\dfrac{\dfrac{4^{n+1}}{(n + 1)^{1}}}{\dfrac{4^{n}}{n!}}]\) Any hints on what is going on here?

\(\displaystyle \dfrac{\dfrac{4^{n+1}}{(n + 1)!}}{\dfrac{4^{n}}{n!}}=\dfrac{4^{n+1}}{4^n} \dfrac{n!}{(n+1)!}=\dfrac{4}{n+1}\)

 

[Jason76]

\(\displaystyle \lim n \rightarrow \infty[\dfrac{\dfrac{4^{n+1}}{(n + 1)^{1}}}{\dfrac{4^{n}}{n!}}]\)

Shouldn't the division process be: \(\displaystyle \dfrac{4^{n+1}}{n+1}(\dfrac{n!}{4^{n}})\)

 

[pka]

Shouldn't the division process be: \(\displaystyle \dfrac{4^{n+1}}{n+1}(\dfrac{n!}{4^{n}})\)

Why do you have \(\displaystyle (n+1)^1\)? It should be \(\displaystyle (n+1)!\)

 

[HallsofIvy]

\(\displaystyle \lim n \rightarrow \infty[\dfrac{a_{n + 1}}{a_{n}}] = L\) Applying this formula to Summation sign with \(\displaystyle \infty\) on top and \(\displaystyle n = 1\) on the bottom \(\displaystyle \dfrac{4^{n}}{n!}\) \(\displaystyle \lim n \rightarrow \infty[\dfrac{\dfrac{4^{n+1}}{(n + 1)^{1}}}{\dfrac{4^{n}}{n!}}]\) Any hints on what is going on here?

What you have s incorrect. That [\(\displaystyle (n+ 1)^1\) should be \(\displaystyle (n+ 1)!\). If \(\displaystyle a_n= \frac{4^n}{n!}\) then \(\displaystyle a_{n+1}= \frac{4^{n+1}}{(n+1)!}= \frac{4(4^n)}{(n+1)n!}\) so the ratio was \(\displaystyle \frac{a_{n+1}}{a_n}=\dfrac{\frac{4(4^n)}{(n+1)n!}}{\frac{4^n}{n!}}\)\(\displaystyle = \dfrac{4(4^n)}{(n+1)n!}\dfrac{n!}{4^n}=\dfrac{4}{n+}\)

 

[Jason76]

How does \(\displaystyle a_{n}\) become \(\displaystyle \dfrac{4^{n}}{n!}\)

or how does \(\displaystyle a_{n+1}\) become \(\displaystyle \dfrac{4^{n+1}}{(n + 1)!}\)

I kind of see a pattern here. Whatever the sub is, then that becomes a power in the numerator and a factorial in the denominator, and the a becomes whatever the target number is. But what is the actual math rule that is being applied here?

\(\displaystyle \dfrac{4^{n+1}}{(n + 1)!} = \dfrac{4(4^{n})}{(n + 1)n!}\) What's going on with this line?

 

[Jason76]

pre-cal was a summer course, so I will do some review on my own. The thing with \(\displaystyle 4^{n + 1}\) was a decomposition expression. You would have to spot those, which can be tricky. Usually your going from simple to complex, not visa versa. Otherwise everything else written makes sense (except for a few things). So now we have:

\(\displaystyle \lim n \rightarrow \infty \dfrac{4}{n + 1} = 0\) so \(\displaystyle 0 < 1\) therefore this thing converges.

 

[Jason76]

Again, though, I don't understand

\(\displaystyle a_{n}\) becoming \(\displaystyle \dfrac{a^{n}}{n!}\) and

\(\displaystyle a_{n + 1}\) becoming \(\displaystyle \dfrac{a^{n+1}}{(n + 1)!}\)

I see a pattern going on, but don't know the reasoning behind it.

 

[daon2]

Again, though, I don't understand

\(\displaystyle a_{n}\) becoming \(\displaystyle \dfrac{4^{n}}{n!}\) and

(replaced a with 4). This is the definition of \(\displaystyle a_n\). It doesn't "become" \(\displaystyle a_n\).

\(\displaystyle a_{n + 1}\) becoming \(\displaystyle \dfrac{4^{n+1}}{(n + 1)!}\)

You get this by replacing every n you see with an (n+1).