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 ?
$\endgroup$ 62 Answers
$\begingroup$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.
$\endgroup$ 9 $\begingroup$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 $y=e^{ln(3)}=3$.
I think is it correct,if wrong tell me and i will change it
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