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$\lim_{n\to\infty}((\frac94)^n+(1+\frac1n)^{n^2})^{\frac1n}$

Here's what I did:

$\lim_{n\to\infty}((\frac94)^n+(1+\frac1n)^{n^2})^{\frac1n}\\ =(\lim(\frac94)^n+\lim((1+\frac1n)^{n})^n)^{\frac1n}\\ =(\lim(\frac94)^n+\lim e^n)^{\frac1n}\\$

Any hints on how to continue?

PS: no logs/integration/derivation because we haven't covered it.

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3  
You can't push the limit inside. –  egreg Nov 29 '13 at 21:57
1  
I think the first step might actually be to recombine: $$1+\frac 1n=\frac {1+n}n$$ then the next step is $$\left(\frac 94\right)^n+\left(\frac{n+1}n\right)^{n^2}=\frac {(9n^n)^n+(n+1)^{n^2}}{(4n^n)^n}$$ –  abiessu Nov 29 '13 at 22:00
    
@abiessu, souldn't that be: $$\frac {(9n^n)^n+4^n(n+1)^{n^2}}{(4n^n)^n}$$ ? Also, I can't find how does that help... –  GinKin Nov 29 '13 at 22:21
2  
This question begs for logarithms and derivatives... Why restrict yourself? –  Chris K Nov 29 '13 at 22:29
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Well... you haven't done l'Hopital's Rule? Or logarithms? –  Chris K Nov 29 '13 at 22:31

4 Answers 4

up vote 7 down vote accepted

$$ \begin{align} &\lim_{n\to\infty}\left(\left(\frac94\right)^n+\left(1+\frac1n\right)^{n^2}\right)^{1/n}\tag{1}\\ &=\lim_{n\to\infty}\left(1+\frac1n\right)^n\lim_{n\to\infty}\left(\left(\frac94\left(1+\frac1n\right)^{-n}\right)^n+1\right)^{1/n}\tag{2}\\ &=e\lim_{n\to\infty}\left(\left(\frac94e^{-1}\right)^n+1\right)^{1/n}\tag{3}\\[9pt] &=e\lim_{n\to\infty}\left(0+1\right)^{1/n}\tag{4}\\[18pt] &=e\tag{5} \end{align} $$ Explanation:

$(1)$: original expression
$(2)$: bring a factor of $\left(1+\frac1n\right)^n$ outside the parentheses
$(3)$: evaluate $\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n=e$
$(4)$: evaluate $\lim\limits_{n\to\infty}\left(\frac94e^{-1}\right)^n=0$
$(5)$: evaluate $\lim\limits_{n\to\infty}1^{1/n}=1$

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Please explain what and how did you do this: $ \lim_{n\to\infty}(1+\frac1n)^n \lim_{n\to\infty} (\left(\frac94(1+\frac1n)^{-n}\right)^n+1)^{1/n} $ –  GinKin Nov 29 '13 at 23:11
    
@GinKin: factored $\left(1+\frac1n\right)^n$ out of the expression and put it on the left. That means dividing everything on the inside of the parentheses by $\left(1+\frac1n\right)^{n^2}$. –  robjohn Nov 29 '13 at 23:22
    
It's basically like Matik Ken's answer. What makes me confused is PEMDAS, would you still be able to factor out even if it wasn't the $nth$ root i.e if it was to the power of n instead ? –  GinKin Nov 29 '13 at 23:28
    
@GinKin: step $(2)$ is only algebra. Even if the limits were not there, step $(2)$ would be valid. –  robjohn Nov 29 '13 at 23:32
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@ParamanandSingh: We can apply that if the family $\{f_n\}$ is equicontinuous at $x_\infty=\lim\limits_{n\to\infty}x_n$ , then $\lim\limits_{n\to\infty}f_n(x_n)=\lim\limits_{n\to\infty}f_n(x_\infty)$. $$ \left(\left(\frac94x\right)^n+1\right)^{1/n}\text{ is equicontinuous at }x=e^{-1}\text{ for }(3) $$ and $$ (x+1)^{1/n}\text{ is equicontinuous at }x=0\text{ for }(4) $$ –  robjohn Nov 30 '13 at 4:38

We know that the sequence $(a_n)$ defined by $$ a_n=\left(1+\frac1n\right)^n $$ converges and its limit is $e$. Notice that $$ \left[\left(\frac94\right)^n+\left(1+\frac1n\right)^{n^2}\right]^{1/n}=\left[\left(\frac94\right)^n+a_n^n\right]^{1/n}=a_n\left[1+\left(\frac{9}{4a_n}\right)^n\right]^{1/n}. $$ Since $$ \lim_{n\to\infty}\frac{9}{4a_n}=\frac{9}{4e}<1, $$ it follows that $$ \lim_{n\to\infty}\left[\left(\frac94\right)^n+\left(1+\frac1n\right)^{n^2}\right]^{1/n}=\lim_{n\to\infty}a_n\left[1+\left(\frac{9}{4a_n}\right)^n\right]^{1/n}=e(1+0)^0=e. $$

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The limit is indeed $e$.

Let $a(n) = \frac94$ and $b(n) = (1+\frac1n)^n$ for $n \ge 1$.

Since $\lim b(n) = e > 9/4 = 2.25$ ( versus $2.71728\ldots$ ) there is an $N$ such that if $n \ge N$ then $b(n) > 9/4$. Thus $x(n)= \frac{a(n)}{b(n)} < 1$ for $n \ge N$.

Let $S(n) =$ the original expression, then:

$S(n) = (a(n)^n + b(n)^n)^{\frac1n} = b(n)(1+(\frac{a(n)}{b(n)})^n)^{\frac1n} = b(n)r(n)$

whereas $r(n) = (1+(\frac{a(n)}{b(n)})^n)^{\frac1n}$ .

We now estimate $r(n)$. It is easy to see that $1 < r(n) < 2^{\frac1n}$ for $n \ge N$.

This means $\lim r(n) = 1$ and so $ \lim S(n) = e$ as claimed.

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For positive $n$, we have that $0 < \left(1 + \frac{1}{n}\right)^{n^2} = \left(\left(1 + \frac{1}{n}\right)^n\right)^n < e^n$. Thus, we can write $$\left(1 + \frac{1}{n}\right)^{n^2} < \left(\frac{9}{4}\right)^n + \left(1 + \frac{1}{n}\right)^{n^2} < \left(\frac{9}{4}\right)^n + e^n < 2 e^n$$ $$\left(\left(1 + \frac{1}{n}\right)^{n^2}\right)^{1/n} < \left(\left(\frac{9}{4}\right)^n + \left(1 + \frac{1}{n}\right)^{n^2}\right)^{1/n} < \left(2e^n\right)^{1/n}.$$

We can use the squeeze theorem on the last inequality to obtain the limit.

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Um correct me if I'm wrong but using the squeeze theorem on this inequality will yield that the limit is 9/4 but from W|A I know that the limit is e. –  GinKin Nov 29 '13 at 22:19
    
You're right: for some reason, when I was writing the answer, I was convinced $\frac{9}{4} = 2.75 > e$. I fixed that, so now it should be right. –  Strants Nov 29 '13 at 22:36
    
Well it seems like you just arbitrarily choose one of the two expression and apply the squeeze theorem to it. What would have happened if the expressions were more complex so you wouldn't know which one was bigger ? Also, how do you know that what you did last time was wrong without checking with W|A ? Sorry if I don't understand enough but it seems like guesswork. –  GinKin Nov 29 '13 at 22:41
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Well, first off, I decided I would try the squeeze theorem, because if $a(n) < b(n) < c(n)$, the $(a(n))^{\frac{1}{n}} < (b(n))^{\frac{1}{n}} < (c(n))^{\frac{1}{n}}$, so I can just ignore the $\frac{1}{n}$ exponent for a while, then add it back in at the end. From there, I knew that $\left(1 + \frac{1}{n}\right)^n < e$, so $\left(1 + \frac{1}{n}\right)^{n^n} < e^n$. Now, in the expression $\left(\frac{9}{4}\right)^n + e^n$, $e^n$ will dominate for large $n$ (since $e \approx 2.71 > 2.25 = \frac{9}{4}$), so I decided to eliminate the less important $\left(\frac{9}{4}\right)^n$ term. –  Strants Nov 30 '13 at 0:02

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