# Math Help - exponential limit

1. ## exponential limit

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

2. What does n approach?.

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

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

Then, $\lim_{t\to 0}\left(1+\frac{3}{t}\right)^{t}=1$

3. Originally Posted by winterwyrm
ok, the lim of the nth root of (1+3n) Thanks in advance!
By the connection between the root and ratio test

$\lim_{n\to\infty}(1+3n)^{\frac{1}{n}}=\lim_{n\to\i nfty}\frac{1+(3n+1)}{1+3n}=1$

4. Originally Posted by Mathstud28
By the connection between the root and ratio test

$\lim_{n\to\infty}(1+3n)^{\frac{1}{n}}=\lim_{n\to\i nfty}\frac{1+(3n+1)}{1+3n}=1$
Que est?

5. Originally Posted by mr fantastic
Que est?
"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

Now, if you want to prove it, that's another matter lol

6. Originally Posted by mr fantastic
Que est?
Qu-est que tu veux?

7. Originally Posted by Moo
"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

Now, if you want to prove it, that's another matter lol
But where is the power series here? Is it meant to be $\sum (1 + 3n) x^n$ .....?

Originally Posted by Mathstud28
By the connection between the root and ratio test

If $a_n = 1 + 3n$ 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 $\sum_{n=1}^{\infty} 2^{(-1)^n - n}$ .....?

8. Originally Posted by Moo
Now, if you want to prove it, that's another matter lol
The proof is not that bad.

9. Originally Posted by mr fantastic
But where is the power series here? Is it meant to be $\sum (1 + 3n) x^n$ .....?
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

@ TPH : would you mind posting it ? ^^

10. Originally Posted by Moo
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

@ TPH : would you mind posting it ? ^^
I understand what you are saying and fully agree (I'm not that senile - actually I knew this already. Hmmmppph .... learned it no more than 80 or 90 years ago).

But what Mathstud said is wrong in the context he used it in. (Otherwise I have completely misunderstood).

11. Originally Posted by mr fantastic
I understand what you are saying and fully agree (I'm not that senile - actually I knew this already. Hmmmppph .... learned it no more than 80 or 90 years ago).

But what Mathstud said is wrong in the context he used it in. (Otherwise I have completely misunderstood).
Well, Mr. F, in what context is that?

12. Originally Posted by Mathstud28
Well, Mr. F, in what context is that?
Originally Posted by Mathstud28
By the connection between the root and ratio test
*Ahem*

Originally Posted by Mr Fantastic
[snip]
What happens when you apply the ratio test and the nth root test to $\sum_{n=1}^{\infty} 2^{(-1)^n - n}$ .....?
Since one test gives 1/2 and the other gives 1, how do you justify your approach to the limit posed by the OP ...... ?

13. Originally Posted by mr fantastic
*Ahem*

Since one test gives 1/2 and the other gives 1, how do you justify your approach to the limit posed by the OP ...... ?
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

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

And $f(n)$ cannot have a variable exponent. I will post back later with a more in-depth reasoning.

14. Originally Posted by Mathstud28
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

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

And $f(n)$ cannot have a variable exponent. I will post back later with a more in-depth reasoning.