Question

For the following sequence, determine the convergence or divergence. If it converges, find the limit.

Answer 1

\[\left\{ a _{n} \right\} = \left\{ \sqrt[n]{n} \right\}\]

Answer 2

type faster girl

Answer 3

help? :(

Answer 4

lol

Answer 5

you are looking for
\[\lim_{n\to \infty}\sqrt[n]{n}\] or
\[\lim_{n\to \infty}n^{\frac{1}{n}}\] which is of the form \(\infty^0\)
use calculus (l'hopital) to get it

Answer 6

oh i see!

Answer 7

you remember how to do it?

Answer 8

i would let f(x)=x^(1/x)

Answer 9

if you like

Answer 10

i need to introduce ln to the equation right?

Answer 11

yes that is one way
the other is to write
\[\large x^{\frac{1}{x}}=e^{\frac{\ln(x)}{x}}\] and compute the limit in the numerator
the form is different, but the work is all the same, computing
\[\lim_{x\to \infty}\frac{\ln(x)}{x}\]

Answer 12

i meant "compute the limit in the EXPONENT" not numerator

Answer 13

yes we weren't shown that in class so i don't know if he would like us to do it that way. in out notes, the prof introduces ln and writes as a fraction to get a fraction of 0/0 so he can apply l'hopital rule

Answer 14

or if you do it your way you get
\[\frac{1}{x}\ln(x)\] with is the same thing
after you get the limit, which is clearly 0 because \(x\) grows way faster than \(\ln(x)\) you say \(e^0=1\)

Answer 15

yes that is the same
as you said "introduce ln' or "take the log" and get
\[\ln(x^{\frac{1}{x}})\] which by the property of the log is the same as
\[\frac{1}{x}\ln(x)=\frac{\ln(x)}{x}\]

Answer 16

so then i take the limit of ln x/x?

Answer 17

now it looks like \(\frac{\infty}{\infty}\) and you can use l'hopital if you must

Answer 18

yes

Answer 19

so for 'hopital rule, i would get the derivatives of the top and bottom?

Answer 20

you don't really need l'hopital because the log grows slower than \(x\) to any positive power

Answer 21

yes, take the derivative top and bottom

Answer 22

so lim x->infinity of ln x/x = 0?

Answer 23

yes

Answer 24

so then it converges using the limit of the function.

Answer 25

if you use l'hopital you see it in one step
you get
\[\frac{\frac{1}{x}}{1}=\frac{1}{x}\] and that limit is clearly 0

Answer 26

don't forget that you took the log as the first step, so now you have to undo that step

Answer 27

\[\lim_{x\to \infty}\frac{\ln(x)}{x}=0\] and so
\[\lim_{x\to \infty}x^{\frac{1}{x}}=e^0\]

Answer 28

which is one?

Answer 29

not sure what you are asking

Answer 30

so my lim x->∞ f(x)= lim x->∞ e^ln f(x) = e^ lim x->∞ ln f(x) = e^0 = 1?

Answer 31

yes

Answer 32

which means it converges by the limit of the function?

Answer 33

i have another question if you dont mind!