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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 ?

Nicholas Stull
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Ignace
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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 07 '12 at 11:43
  • Take some values of n say like n = 1 to 20, then evaluate $(2^n + 3^n)^{\frac{1}{n}}$. What do you observe? – TenaliRaman May 07 '12 at 11:50
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    @Ignace I tried to edit as much as I could. (There was no latex at all in your post). – Kirthi Raman May 07 '12 at 11:50
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    @Ignace, Didier pretty much gave you a "big hint" and you don't need anything more to solve this. – Kirthi Raman May 07 '12 at 11:57
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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. – André Nicolas May 07 '12 at 13:13
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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. – MJD May 07 '12 at 13:26

2 Answers2

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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.

Simon Markett
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  • is not limit infinity?if we take logarithm of both side and use L'hopital rule? – dato datuashvili May 07 '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 07 '12 at 12:44
  • no you are right,i forgot ln while calculation derivative – dato datuashvili May 07 '12 at 12:52
  • i got answer of this limit=3 – dato datuashvili May 07 '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 08 '12 at 14:37
  • The limit is 0. Use the inequality $n!\leq (\frac{n+1}{2})^n$. To see that this inequality holds you only need $(n-x)(n+x)\leq n^2$. – Simon Markett May 08 '12 at 15:01
  • I don't see the point. Is the first inequality a known one. And what do the x's in the second do to prove the first. And does it all prove the given problem n!/n^n. – Ignace May 08 '12 at 21:16
  • So you ask me to solve it for you completely? Did you try working with my hints? I am saying 1. If you know the inequality it follows quite easily that your sequence converges to 0 and 2. The inequality is not hard to prove. Obviously there are many ways to do this. One possibility is to take two factors of $n!$ at a time and compare them to a certain square. – Simon Markett May 09 '12 at 13:04
  • No, I would like to find out myself. But any books on the subject I have fail to clear out the basics for me. Any advise on a solid introduction to the subject, with good examples would be welcome. – Ignace May 10 '12 at 08:11
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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

Thomas
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dato datuashvili
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