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It is known that for integers $n \geq 1$,

$$\lim_{ n \to \infty} (1 + n)^{1/n} = e = 2.718\dots$$

For integer $N \ge n$, is it true that:

$$\lim_{ n, N \to \infty} (1 + n + N)^{1/n} > e\ \ ?$$

Suppose the sequence is monotone either increasing or decreasing and also that infinitely many terms of the limit

$$\lim_{ n, N \to \infty} (1 + n + N)^{1/n}$$

are bounded within some compact interval $[a, b]$ on the real line. Is this limit

$$\lim_{ n, N \to \infty }(1 + n + N)^{1/n}$$

finite on $[a, b]$? Does it converge to some finite value?

If anyone can help to solve this question then I thank you in advance.

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2  
The answers depend on whether you take the limits with respect to $n$ or $N$ first. (If you do both at the same time the limit won't exist, because it would require the two differently-ordered limits to be equivalent at least.) –  anon Aug 25 '12 at 21:04
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The limit you gave has value $1$, not $e$. You want $(1+1/n)^n$. –  André Nicolas Aug 25 '12 at 21:15
    
I think he really wants $(1+n)^{1/n}$, but with $n\rightarrow 0$. Maybe not...i dont know. –  Integral Aug 25 '12 at 21:58
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You originally had $n=\gt1$ and $N=\gt n$; different people changed this to $n\ge1$ and $N\to n$, respectively; it seems unlikely that this was what you intended. I've now changed $N\to n$ to $N\ge n$; please check whether this is what you wanted. To avoid all this confusion in the future, please use $\TeX$ to format your posts yourself. Inline formulas are enclosed in single dollar signs, displayed equations in double dollar signs; you can get the code for any math formatting you see on this site by right-clicking on it and selecting "Show Math As:TeX Commands". –  joriki Aug 26 '12 at 9:28

1 Answer 1

Your question makes very little sense, especially the part about the compact interval $[a,b]$ - what does that have to do with $n$ and $N$, which it seems are both going to infinity?

Anyway, $$\lim_{n\to\infty}\lim_{N\to\infty}(1+n+N)^{1/n}$$ doesn't exist (or, if you prefer, is infinite), while $$\lim_{N\to\infty}\lim_{n\to\infty}(1+n+N)^{1/n}=1$$ If neither of these is what you want, please edit your question to clarify.

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1  
As already mentioned in the comments, the second limit in your post is NOT exp(1). –  Did Aug 26 '12 at 7:57
    
@did, for fixed $N$, $(1+n+N)^{1/n}=(1+N)^{1/n}(1+{n\over1+N})^{1/n}$. So $(1+N)^{1/n}\to1$ and $(1+{n\over1+N})^{1/n}\to e^{1/(1+N)}$. Then take the limit on $N$ and get $e$. What'd I do wrong? –  Gerry Myerson Aug 26 '12 at 11:02
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Note that $(1+an)^{1/n}$ does not converge to $\mathrm e^a$ when $n\to+\infty$, for a fixed $a\gt0$ (taking logarithms, one sees that the limit is $1$). The limit $(1+ax)^{1/x}\to\mathrm e^a$ is valid when $x\to0$. –  Did Aug 26 '12 at 11:33
    
@did, I hate it when I do that. I've made an edit. –  Gerry Myerson Aug 26 '12 at 12:52
    
Gerry: Relax, I suffered briefly from the same hallucination but @André's comment saved me... :-) –  Did Aug 26 '12 at 12:57

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