Root Test for Divergence or Convergence - Problem # 2

[Jason76]

Root test for divergence or convergence of a series.

If \(\displaystyle L < 0\) then it converges

If \(\displaystyle L >0\) then it diverges

If \(\displaystyle L = 1\) then the problem does not provide us with the needed info.

Summation sign with \(\displaystyle \infty\) on top and \(\displaystyle n = 1\) on the bottom \(\displaystyle \lim n \rightarrow \infty \dfrac{e^{3n}}{n^{n}}\)

\(\displaystyle \lim n \rightarrow \infty \dfrac{e^{3n}}{n^{n}}\)

\(\displaystyle \lim n \rightarrow \infty [\dfrac{e^{3n}}{n^{n}}]^{ln}\)

\(\displaystyle \lim n \rightarrow \infty \dfrac{e^{3}}{n}\) How did this line become like this?

\(\displaystyle \lim n \rightarrow \infty\dfrac{\dfrac{e^{3}}{n}}{\dfrac{n}{n}}\)

\(\displaystyle \lim n \rightarrow \infty\dfrac{\dfrac{e^{3}}{\infty}}{1}\)

\(\displaystyle \lim n \rightarrow \infty\dfrac{0}{1} = 0\)

\(\displaystyle 0 < 1\) so it converges

 

[Romsek]

Root test for divergence or convergence of a series.

If \(\displaystyle L < 0\) then it converges

If \(\displaystyle L >0\) then it diverges

If \(\displaystyle L = 1\) then the problem does not provide us with the needed info.

Summation sign with \(\displaystyle \infty\) on top and \(\displaystyle n = 1\) on the bottom \(\displaystyle \lim n \rightarrow \infty \dfrac{e^{3n}}{n^{n}}\)

\(\displaystyle \lim n \rightarrow \infty \dfrac{e^{3n}}{n^{n}}\)

\(\displaystyle \lim n \rightarrow \infty [\dfrac{e^{3n}}{n^{n}}]^{ln}\)

\(\displaystyle \lim n \rightarrow \infty \dfrac{e^{3}}{n}\) How did this line become like this?

it should be \(\displaystyle \left(\frac{e^3}{n}\right)^n\)

and then take this to the power 1/n for the root test.

 

[Jason76]

I caught the error after reviewing the vids. \(\displaystyle \dfrac{1}{n}\) looked like \(\displaystyle \ln\)

Root test for divergence or convergence of a series.

If \(\displaystyle L < 0\) then it converges

If \(\displaystyle L >0\) then it diverges

If \(\displaystyle L = 1\) then the problem does not provide us with the needed info.

Summation sign with \(\displaystyle \infty\) on top and \(\displaystyle n = 1\) on the bottom \(\displaystyle \lim n \rightarrow \infty \dfrac{e^{3n}}{n^{n}}\)

\(\displaystyle \lim n \rightarrow \infty [\dfrac{e^{3n}}{n^{n}}]^{\dfrac{1}{n}}\)

\(\displaystyle \lim n \rightarrow \infty \dfrac{e^{3}}{n}\)

\(\displaystyle \lim n \rightarrow \infty\dfrac{\dfrac{e^{3}}{n}}{\dfrac{n}{n}}\)

\(\displaystyle \lim n \rightarrow \infty\dfrac{\dfrac{e^{3}}{\infty}}{1}\)

\(\displaystyle \lim n \rightarrow \infty\dfrac{0}{1} = 0\)

\(\displaystyle 0 < 1\) so it converges