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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 $u^n\to0$.2012-05-07
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    Take some values of n say like n = 1 to 20, then evaluate $(2^n + 3^n)^{\frac{1}{n}}$. What do you observe?2012-05-07
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    @Ignace I tried to edit as much as I could. (There was no latex at all in your post).2012-05-07
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    @Ignace, Didier pretty much gave you a "big hint" and you don't need anything more to solve this.2012-05-07
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    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.2012-05-07
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    $\infty^0$ 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.2012-05-07

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