Root Test for Divergance or Convergance

[Jason76]

Does this diverge or converge based on root test?

If \(\displaystyle L < 1\) then it converges, if \(\displaystyle L > 1\) then it diverges. If \(\displaystyle L = 1\) then the test cannot provide the needed info.

Summation symbol with \(\displaystyle \infty\) on top of the symbol and \(\displaystyle n = 1\) on the bottom \(\displaystyle \dfrac{n^{n - 1}}{2^{3+n}}\) (line from online video I copied)

\(\displaystyle \lim x \rightarrow \infty \dfrac{n^{n - 1}}{2^{3+n}}\) - (line from online video I copied)

\(\displaystyle \lim x \rightarrow \infty [\dfrac{n^{n - 1}}{2^{3+n}}]^{ln}\) (line from online video I copied)

\(\displaystyle \lim x \rightarrow \infty \dfrac{(n^{n - 1})^{ln}}{(2^{3+n})^{ln}}\)

What next? I know the \(\displaystyle \ln\) of 1 is 0

Is \(\displaystyle \lim x \rightarrow \infty [\dfrac{n^{n - 1}}{2^{3+n}}]^{ln}\) the same as \(\displaystyle \lim x \rightarrow \infty \ln[\dfrac{n^{n - 1}}{2^{3+n}}]\) which would lead to
\(\displaystyle \lim x \rightarrow \infty \dfrac{\ln (n^{n - 1})}{\ln (2^{3+n})}\)

 

[pka]

Does this diverge or converge based on root test?
If \(\displaystyle L < 1\) then it converges, if \(\displaystyle L > 1\) then it diverges. If \(\displaystyle L = 1\) then the test cannot provide the needed info.
Summation symbol with \(\displaystyle \infty\) on top of the symbol and \(\displaystyle n = 1\) on the bottom \(\displaystyle \dfrac{n^{n - 1}}{2^{3+n}}\) (line from online video I copied)
\(\displaystyle \lim x \rightarrow \infty \dfrac{n^{n - 1}}{2^{3+n}}\) - (line from online video I copied)
\(\displaystyle \lim x \rightarrow \infty [\dfrac{n^{n - 1}}{2^{3+n}}]^{ln}\) (line from online video I copied)
\(\displaystyle \lim x \rightarrow \infty \dfrac{(n^{n - 1})^{ln}}{(2^{3+n})^{ln}}\)
What next? I know the \(\displaystyle \ln\) of 1 is 0
Is \(\displaystyle \lim x \rightarrow \infty [\dfrac{n^{n - 1}}{2^{3+n}}]^{ln}\) the same as \(\displaystyle \lim x \rightarrow \infty \ln[\dfrac{n^{n - 1}}{2^{3+n}}]\) which would lead to
\(\displaystyle \lim x \rightarrow \infty \dfrac{\ln (n^{n - 1})}{\ln (2^{3+n})}\)

@Jason76, That may well be the most irresponsible post you have ever made.
We have absolutely no idea what question you are answering. What is the question?

You have put us down in the middle of something going on in your confused brain.

 

[Jason76]

Does this diverge or converge based on root test?

If \(\displaystyle L < 1\) then it converges, if \(\displaystyle L > 1\) then it diverges. If \(\displaystyle L = 1\) then the test cannot provide the needed info.

Summation symbol with \(\displaystyle \infty\) on top of the symbol and \(\displaystyle n = 1\) on the bottom \(\displaystyle \dfrac{n^{n - 1}}{2^{3+n}}\) (line from online video I copied)

\(\displaystyle \lim x \rightarrow \infty \dfrac{n^{n - 1}}{2^{3+n}}\) - (line from online video I copied)

\(\displaystyle \lim x \rightarrow \infty [\dfrac{n^{n - 1}}{2^{3+n}}]^{ln}\) (line from online video I copied)

What is the next move?

 

[Jason76]

Does this diverge or converge based on root test?

If \(\displaystyle L < 1\) then it converges, if \(\displaystyle L > 1\) then it diverges. If \(\displaystyle L = 1\) then the test cannot provide the needed info.

Summation symbol with \(\displaystyle \infty\) on top of the symbol and \(\displaystyle n = 1\) on the bottom \(\displaystyle \dfrac{n^{n - 1}}{2^{3+n}}\) (line from online video I copied)

\(\displaystyle \lim x \rightarrow \infty \dfrac{n^{n - 1}}{2^{3+n}}\)

\(\displaystyle \lim x \rightarrow \infty [\dfrac{n^{n - 1}}{2^{3+n}}]^{ln}\) How do I deal with the \(\displaystyle \ln\) here?

 

[daon2]

Does this diverge or converge based on root test?

If \(\displaystyle L < 1\) then it converges, if \(\displaystyle L > 1\) then it diverges. If \(\displaystyle L = 1\) then the test cannot provide the needed info.

Summation symbol with \(\displaystyle \infty\) on top of the symbol and \(\displaystyle n = 1\) on the bottom \(\displaystyle \dfrac{n^{n - 1}}{2^{3+n}}\) (line from online video I copied)

\(\displaystyle \lim x \rightarrow \infty \dfrac{n^{n - 1}}{2^{3+n}}\)

\(\displaystyle \lim x \rightarrow \infty [\dfrac{n^{n - 1}}{2^{3+n}}]^{ln}\) How do I deal with the \(\displaystyle \ln\) here?

Writing "ln" as a power is nonsense to me, I have no idea what you're doing.

 

[Jason76]

Does this diverge or converge based on root test?

If \(\displaystyle L < 1\) then it converges, if \(\displaystyle L > 1\) then it diverges. If \(\displaystyle L = 1\) then the test cannot provide the needed info.

Summation symbol with \(\displaystyle \infty\) on top of the symbol and \(\displaystyle n = 1\) on the bottom \(\displaystyle \dfrac{n^{n - 1}}{2^{3+n}}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{n^{n - 1}}{2^{3+n}}\)

\(\displaystyle \lim x \rightarrow \infty [\dfrac{n^{n - 1}}{2^{3+n}}]^{\dfrac{1}{n}}\) changed line

 

[daon2]

Does this diverge or converge based on root test?

If \(\displaystyle L < 1\) then it converges, if \(\displaystyle L > 1\) then it diverges. If \(\displaystyle L = 1\) then the test cannot provide the needed info.

Summation symbol with \(\displaystyle \infty\) on top of the symbol and \(\displaystyle n = 1\) on the bottom \(\displaystyle \dfrac{n^{n - 1}}{2^{3+n}}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{n^{n - 1}}{2^{3+n}}\)

\(\displaystyle \lim x \rightarrow \infty [\dfrac{n^{n - 1}}{2^{3+n}}]^{\dfrac{1}{n}}\) changed line

\(\displaystyle \left(\dfrac{n^{n-1}}{2^{n+3}}\right)^{1/n} =\dfrac{1}{(8n)^{1/n}}\left(\dfrac{n^n}{2^{n}}\right)^{1/n}=\dfrac{1}{(8n)^{1/n}}\cdot \dfrac{n}{2}\)

Then notice \(\displaystyle (8n)^{1/n} \le 8\sqrt{n}\)

 

[stapel]

Does this diverge or converge based on root test?

We have absolutely no idea what question you are answering. What is the question?

Does this diverge or converge based on root test?

Writing "ln" as a power is nonsense to me, I have no idea what you're doing.

Does this diverge or converge based on root test?

At some point, it would be really nice if you would answer questions, rather than merely repeating cryptic original statements that are clearly not being understood.

Seriously, we really can't read your mind.

 

[HallsofIvy]

Does this diverge or converge based on root test?

If \(\displaystyle L < 1\) then it converges, if \(\displaystyle L > 1\) then it diverges. If \(\displaystyle L = 1\) then the test cannot provide the needed info.

Summation symbol with \(\displaystyle \infty\) on top of the symbol and \(\displaystyle n = 1\) on the bottom \(\displaystyle \dfrac{n^{n - 1}}{2^{3+n}}\) (line from online video I copied)

\(\displaystyle \lim x \rightarrow \infty \dfrac{n^{n - 1}}{2^{3+n}}\) - (line from online video I copied)

\(\displaystyle \lim x \rightarrow \infty [\dfrac{n^{n - 1}}{2^{3+n}}]^{ln}\) (line from online video I copied)

What is this "ln" exponent? You mean "1/n", the nth root, don't you? This is, after all, the "root test"! You are copying things without having any idea what they mean!

\(\displaystyle \lim x \rightarrow \infty \dfrac{(n^{n - 1})^{ln}}{(2^{3+n})^{ln}}\)

What next? I know the \(\displaystyle \ln\) of 1 is 0

But you don't have "ln". Even if you thought the "1/n" really was "ln", the logarithm is a function, not a number. You must have "ln" of something.

Is \(\displaystyle \lim x \rightarrow \infty [\dfrac{n^{n - 1}}{2^{3+n}}]^{ln}\) the same as \(\displaystyle \lim x \rightarrow \infty \ln[\dfrac{n^{n - 1}}{2^{3+n}}]\) which would lead to
\(\displaystyle \lim x \rightarrow \infty \dfrac{\ln (n^{n - 1})}{\ln (2^{3+n})}\)

 

[Jason76]

What is this "ln" exponent? You mean "1/n", the nth root, don't you? This is, after all, the "root test"! You are copying things without having any idea what they mean!


But you don't have "ln". Even if you thought the "1/n" really was "ln", the logarithm is a function, not a number. You must have "ln" of something.

I understand that a \(\displaystyle \ln\) cannot be used as an exponent, that's why I was confused when I saw it as an exponent. But I was simply misreading the vid in which the professor wrote \(\displaystyle \dfrac{1}{n}\) and it looked like natural log.