Question

I'm just bad at the properties of exponents and am trying to understand how this works out in relation to series that converge or diverge (nth root of n). Problem posted below in a minute.

Answer 1

If I had a term like

\[\frac{n^{2}}{2^{n}}\]
and I applied the root test to it as so:

\[\sqrt[n]{\frac{n^{2}}{2^{n}}}\]
Why would, after simpifying the denominator to 2, the numerator end up being 1^2?

I'm having an issue sorting out proper and improper ways to express the denominator (cont'd)

Answer 2

So would this be a valid way?

Just dealing with the denominator on its own:

\[\sqrt[n]{n ^{2}} = n ^{2}*n ^{ \frac{1}{n}}\]

Answer 3

Or \[n^{2 + \frac{1}{n}}\]? The last one is what I'm really concerned with. Are these three all valid ways of expressing the exponent?

Answer 4

(It is assumed knowledge for this problem that lim (n -> infinity) of the nth root of n converges to 1. I understand that the numerator can be expressed as such (nth root of n in parentheses, squared), so why are these other methods not valid? If the limit is taken while they're in this form, you won't get 1, which should be the answer.)

Answer 5

@Psymon , if you get the chance; I'm just a little confused about how the exponents are being treated; \[\sqrt[n]{n ^{2}} = (\sqrt[n]{n})^{2}\] I'm sure of that. But if that's the case, I don't understand why \[n^{2(1/n)} = n^{\frac{2}{n}}\] isn't also a valid limit; If I took the limit of that, I would get infinity^zero, and my teacher said you have to use L'Hopital's rule to resolve the other way this exponent can be expressed, e.g. this.

Answer 6

Here's my understanding:

\[n^{2}*n^{\frac{1}{n}} = n^{n + \frac{1}{n}}\]
\[\sqrt[n]{n^{2}} = (\sqrt[n]{n})^{2} =n^{2* \frac{1}{n}}\]

Answer 7

Whoops, top right side exponent should be 2 + (1/n).

Answer 8

\[\sqrt[a]{b^c}=b^{\frac{c}{a}} \]

Answer 9

Bottom right exponent is multiplication, just for clarity.

Answer 10

im confused about your question. what exactly are you asking?

Answer 11

In terms of just a limit, x^(2/x) to infinity gets treated just as if it were 2^(2/x)or something, the answer is 1

Answer 12

I'm asking what are the other valid ways of expressing
\[\sqrt[n]{n^{2}},\]
Which I believe to be\[\sqrt[n]{n^{2}} = (\sqrt[n]{n})^{2} = n^{\frac{1}{n}(2)}\]

Answer 13

Psymon: I never knew that, I guess, wow. Thanks. Let me just check on this because it feels like you just told me something I should have been aware of for a while but apparently haven't been.

Answer 14

that is only true for some cases
think of
\[\sqrt[2]{(-2)^2}=\sqrt{4} = 2 \ne \sqrt{-2}^2 \ is \ not \ defined\]

Answer 15

What is only true for some cases, specifically?

Answer 16

Okay, well in any case, I guess my prof. misspoke when he said it would be resolve by L'Hopital if expressed in a different form.

Answer 17

*resolved

Answer 18

\[\lim_{x\rightarrow \infty} \ y = x^{\frac{1}{n}}\\ln(lim_{x\rightarrow \infty}y) = \frac{1}{x}ln(x)\] now with lhopital
\[ln(\lim_{x\rightarrow \infty}y)=\frac{1}{x}\\e^{\lim_{x\rightarrow \infty}y}=e^{0}
\\\lim_{x\rightarrow \infty} y = 1\]

Answer 19

that should say\[ln(\lim_{x\rightarrow \infty}y)=\frac{1}{x}\\e^{ln(\lim_{x\rightarrow \infty}y)}=e^{0}
\\\lim_{x\rightarrow \infty} y = 1\]

Answer 20

Thanks!

Answer 21

np