$\displaystyle a_n= (3^n + 5^n)^{\frac 1n}$

is the above sequence convergent? if so, what is the limit?

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- Mar 14th 2009, 09:59 PMtwilightstr
$\displaystyle a_n= (3^n + 5^n)^{\frac 1n}$

is the above sequence convergent? if so, what is the limit? - Mar 14th 2009, 10:41 PMJhevon
- Mar 14th 2009, 11:10 PMtwilightstr
is it infinity

- Mar 14th 2009, 11:12 PMJhevon
- Mar 14th 2009, 11:18 PMo_O
Just as an alternative..

If you can use the fact that $\displaystyle \lim_{n \to \infty} a^{\frac{1}{n}} = 1$ for $\displaystyle a > 0$, then notice that:

$\displaystyle 5^n \ \ \leq \ \ 3^n + 5^n \ \ \leq \ \ 5^n + 5^n$

$\displaystyle \Rightarrow \left(5^n \right)^{\frac{1}{n}} \ \ \leq \ \ \left(3^n +5^n \right)^{\frac{1}{n}} \ \ \leq \ \ \left(5^n +5^n \right)^{\frac{1}{n}}$

$\displaystyle \Rightarrow 5 \ \ \leq \ \ \left(3^n +5^n \right)^{\frac{1}{n}} \ \ \leq \ \ 5\left(2 \right)^{\frac{1}{n}}$

And use squeeze theorem .. - Mar 14th 2009, 11:23 PMtwilightstr
jhevon, how would u take the limit

- Mar 14th 2009, 11:34 PMJhevon
- Mar 14th 2009, 11:36 PMtwilightstr
yes. thats what i tried doing in the first place, but i think i did it incorrectly.

- Mar 15th 2009, 12:23 AMJhevon
recall that, by L'Hopital's, $\displaystyle \lim_{n \to \infty} \frac {\ln (3^n + 5^n)}n = \lim_{n \to \infty} \frac {\frac d{dn}[ \ln (3^n + 5^n)]}{\frac d{dn}n}$

i suppose it is the $\displaystyle \frac d{dn}[ \ln (3^n + 5^n)]$ that is giving you trouble. what did you get for this? - Mar 15th 2009, 10:19 AMstapel
- Mar 15th 2009, 12:50 PMtwilightstr
= [(ln3)3^x + (ln5)5^x]/(3^x + 5^x) as the limit goes to infinity. I really dont know what to do from there

- Mar 15th 2009, 12:53 PMJhevon
- Mar 15th 2009, 01:05 PMKrizalid
$\displaystyle \left( 3^{n}+5^{n} \right)^{1/n}=5\left\{ \left( \frac{3}{5} \right)^{n}+1 \right\}^{1/n}\to 5$ since $\displaystyle \left( \frac{3}{5} \right)^{n}+1\to 1$ and $\displaystyle 1^{1/n}\to 1$ as $\displaystyle n\to\infty.$