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I was trying to solve the $\lim\limits_{n \to \infty}(1+\frac{1}{n^{3\alpha}})^{n^{5}}$ where I have to say for which $\alpha$ parameter the limit is finite. I tried to sobstitute $t=n^{3\alpha}$:

$(1+\frac{1}{t})^{{t}^{\frac{5}{3\alpha}}}$

and I found $\alpha \ge 0$

But the solution says $\alpha\geq \frac{5}{3}$

Where I'm wrong?

Tarlo_x
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  • You are right, provided this is the right statement of the question. Have you checked what the problem exactly asks? – Clement C. Jun 15 '16 at 19:59
  • Sorry I made some mistakes in writing the question. – Tarlo_x Jun 15 '16 at 20:10
  • That your method lead you to conclude that it would be finite for $\alpha=0$ should by itself make you suspicious: In that case, the initial limit would become $\lim_{n\to\infty} 2^{n^5}$ which definitely doesn't exist. (By contrast, the cases of $\alpha=1$ and $\alpha=2$ are not so obvious and so provide more interesting test cases.) – Semiclassical Jun 15 '16 at 20:25
  • @Tarlo_x Indeed, that makes quite a difference... – Clement C. Jun 15 '16 at 20:28

3 Answers3

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First exclude the case $\alpha\le0$. For $\alpha>0$, compute instead $$ \lim_{x\to\infty}\left(1+\frac{1}{x^{3\alpha}}\right)^{x^5} $$ (the function as opposed to the sequence). As a preparation, we compute the limit of the logarithm, but substituting $x=1/t$: $$ \lim_{t\to0^+}\frac{\log(1+t^{3\alpha})}{t^5}= \lim_{t\to0^+}\frac{t^{3\alpha}+o(t^{3\alpha})}{t^5} $$ and you want the limit is finite. This clearly means $3\alpha\ge 5$.

Fill in the details.

egreg
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  • In my opinion this is by far the simplest, clearest answer. +1 Yet, wouldn't that little "o" there be better expressed as $;\mathcal O(t^{6\alpha});$ = big "O"? – DonAntonio Jun 15 '16 at 20:41
  • @Joanpemo Since the $o(t^{3\alpha})$ is enough, why go further? – Clement C. Jun 15 '16 at 20:52
  • @ClementC. Thank you. That's right, it is just that I've this idea that big O is better well-known and handled than little o...at least in my case. – DonAntonio Jun 15 '16 at 20:59
  • @Joanpemo Maybe big-O is more flexible, but in the context of Taylor expansion usually little-o is sufficient. – egreg Jun 15 '16 at 21:00
  • @Joanpemo Note that here, the statement used is equivalent to $\lim_{u-\to0}\frac{\ln(1+u)}{u} = 1$, i.e. $\ln'(1)=1$. The more general statement is not that basic. – Clement C. Jun 15 '16 at 21:00
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$$\left(1+\frac1{n^{3\alpha}}\right)^{n^5}=\left[\left(1+\frac1{n^{3\alpha}}\right)^{n^{3\alpha}}\right]^{n^{5-3\alpha}}$$

Now, for $\;3\alpha>0\;$ we get $\;n^{3\alpha}\xrightarrow[n\to\infty]{}\infty\;$ , so

$$\left(1+\frac1{n^{3\alpha}}\right)^{n^{3\alpha}}\xrightarrow[n\to\infty]{}e$$, and if $\;5+3\alpha>0\;$ then

$$\left[\left(1+\frac1{n^{3\alpha}}\right)^{n^{3\alpha}}\right]^{n^{5-3\alpha}}\ge 2^{n^{5-3\alpha}}\xrightarrow[n\to\infty]{}\infty$$

so we must have

$$5-3\alpha\le0\iff\alpha\ge\frac53$$

DonAntonio
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1

Starting from what you did:

First of all, for $\alpha\leq 0$ you have $1+\frac{1}{n^{3\alpha}}\geq 2$, and $2^{n^5}$ diverges to $\infty$. This settles this case, so now we consider $\alpha > 0$.

Consider what you have:

$$\begin{align} \left(1+\frac{1}{t}\right)^{{t}^{\frac{5}{3\alpha}}} &= \exp\left({t}^{\frac{5}{3\alpha}} \ln\left(1+\frac{1}{t}\right)\right) = \exp\left(\frac{{t}^{\frac{5}{3\alpha}}}{t}\cdot t \ln\left(1+\frac{1}{t}\right)\right) \\&= \exp\left({t}^{\frac{5-3\alpha}{3\alpha}}\cdot t \ln\left(1+\frac{1}{t}\right)\right) \end{align}$$

Recalling that $t \ln\left(1+\frac{1}{t}\right)\xrightarrow[t\to\infty]{}1$, we get:

  • If $5-3\alpha < 0$, then the exponent goes to $0\cdot 1=0$
  • If $5-3\alpha = 0$, then the exponent goes to $1\cdot 1=1$
  • If $5-3\alpha > 0$, then the exponent goes to $\infty\cdot 1=\infty$

Can you conclude?

Clement C.
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