ok, the lim of the nth root of (1+3n) Thanks in advance!

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- Jun 19th 2008, 02:33 PMwinterwyrmexponential limit
ok, the lim of the nth root of (1+3n) Thanks in advance!

- Jun 19th 2008, 02:37 PMgalactus
What does n approach?.

I assume you mean $\displaystyle \lim_{n\to {\infty}}\left(1+3n\right)^{\frac{1}{n}}$

If so, let $\displaystyle t=\frac{1}{n}$

Then, $\displaystyle \lim_{t\to 0}\left(1+\frac{3}{t}\right)^{t}=1$ - Jun 19th 2008, 07:05 PMMathstud28
- Jun 19th 2008, 07:30 PMmr fantastic
- Jun 19th 2008, 10:23 PMMoo
"Qu'est-ce que c'est ?" ^^

If you want to know, you can remember that when dealing with power series and looking for the radius of convergence, you can choose between the ratio or the power test. Both will give the same result (Tongueout)

Now, if you want to*prove*it, that's another matter lol - Jun 19th 2008, 10:24 PMMathstud28
- Jun 20th 2008, 04:23 AMmr fantastic
But where is the power series here? Is it meant to be $\displaystyle \sum (1 + 3n) x^n$ .....?

Quote:

Originally Posted by**Mathstud28**

If $\displaystyle a_n = 1 + 3n$ then the ratio is wrong for a start.

By the way, regarding "*By the connection between the root and ratio test*" ...... What happens when you apply the ratio test and the nth root test to $\displaystyle \sum_{n=1}^{\infty} 2^{(-1)^n - n}$ .....? (Thinking)

(Tongueout) - Jun 20th 2008, 07:33 AMThePerfectHacker
- Jun 20th 2008, 08:03 AMMoo
Nah, I was talking about when you want to find the radius of convergence of a series, you can use either the ratio test either the root test to reach the

__same__result.

This is a way to "remember" what mathstud used (Tongueout)

@ TPH : would you mind posting it ? ^^ - Jun 20th 2008, 10:58 AMmr fantastic
- Jun 20th 2008, 11:00 AMMathstud28
- Jun 20th 2008, 05:32 PMmr fantastic
- Jun 20th 2008, 07:39 PMMathstud28
Are you really going to be that semantical? Well to answer your question, I never said every limit can be done this way, what you provided is a case where the root test and the ratio test differ. What I am talking about is much more specific, it is

By the connection

$\displaystyle \lim_{n\to\infty}f(n)^{\frac{1}{n}}=\lim_{n\to\inf ty}\frac{f(n+1)}{f(n)}$

And $\displaystyle f(n)$ cannot have a variable exponent. I will post back later with a more in-depth reasoning. - Jun 20th 2008, 08:45 PMmr fantastic