# convergence to infinity of$(2^n + 3^n)^{\frac{1}{n}}$ (as $n \to \infty$)

Just checking. $2^n$ ($n \to \infty$) tends to $\infty$.
+ $3^n$ also ($n \to \infty$) tends to $\infty$

so the sum gets me $\infty$.

Now $(\infty)^{1/\infty}$ : $(\infty)^0 = 1$

I see no other way. Theorem: $n^{\frac{1}{n}} = 1$ as $n \to \infty$.

Analogous: $(2^n - 3^n)^\frac{1}{n}$ = $(\infty - \infty)^0$ = 1 ???

Is it ok to suppose infinity on any integer $k^n$ as $n$ goes to infinity ?

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Hint: this is $3\cdot(1+u^n)^{1/n}$, with $u=2/3\lt1$ hence $u^n\to0$. –  Did May 7 '12 at 11:43
@Ignace I tried to edit as much as I could. (There was no latex at all in your post). –  Kirthi Raman May 7 '12 at 11:50
@Ignace, Didier pretty much gave you a "big hint" and you don't need anything more to solve this. –  Kirthi Raman May 7 '12 at 11:57
General advice: Do not substitute the symbol $\infty$ for $n$ or $x$ in an algebraic expression. In the majority of questions you will be asked, that procedure gives the wrong answer. –  André Nicolas May 7 '12 at 13:13
$\infty^0$ is called an "indeterminate form", and your example is the reason why: when you have a limit that seems at first glance to have this form, you cannot tell what it will do without looking more closely. –  MJD May 7 '12 at 13:26
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Before I answer your question: In general you really should be careful with takink limits seperately. You have probably seen proofs that taking limit behaves well under certain operations if the limit exists, i.e. is finite. When your limits are infinte there might be strange effects. Just look at the series for $e$: $$(1+1/n)^n$$ For expressions like in your example the "sandwich lemma" is off big help. Just observe $$3=(3^n)^{\frac 1n}\leq (2^n+3^n)^{\frac 1n}\leq (2\cdot 3^n)^{\frac 1n}=2^{\frac 1n}\cdot 3\to3$$

For your "analogous" statement: You are aware that the expression in the brackets is negative for all $n\geq1$. Tking the $n^{th}$ root is not really well defined.

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is not limit infinity?if we take logarithm of both side and use L'hopital rule? –  dato datuashvili May 7 '12 at 12:24
@dato I am pretty sure that this solution is correct. If you are not convinced try to explain where my mistake should be. (Sorry if I sound patronising) –  Simon Markett May 7 '12 at 12:44
no you are right,i forgot ln while calculation derivative –  dato datuashvili May 7 '12 at 12:52
i got answer of this limit=3 –  dato datuashvili May 7 '12 at 12:56
Thanks for the ideas. Another that keeps me troubled is n!/n^n. It simplifies to (n-1)!/n^(n-1). Of course the nominator keeps de/or increasing, asthe denominator stays infinite. I'd say the limit is infinite. –  Ignace May 8 '12 at 14:37
We can use following procedure: let $y= (2^n + 3^n)^{\frac{1}{n}}$. Then we can see that $\ln(y)=\frac{\ln(2^n+3^n)}{n}$ on the right side, limit is in the form infinity/infinity, so use L'hopitals rule, take derivatives, we would have $$\frac{2^n\ln(2)+3^n\ln(3)}{2^n+3^n}$$ it's limits is $ln(3)$, so finally we would have $y$ is equal to $e$ in power of $\ln(3)$, so finally $y=e^{ln(3)}=3$.
How did you compute the derivative of $ln(2^x+3^x)$ ? –  Simon Markett May 7 '12 at 12:42