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How do you evaluate this sequence limit using the squeeze/sandwich theorem? $$\lim_{n \to \infty} \left(3^n+1\right)^{\frac1n} $$

I don't really know where to start. I've tried using the fact that $\lim_{n \to \infty} \left(3^n\right)^{\frac1n} = 3$ (which is the correct answer) but I don't know where to go from there.

Thanks!

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7 Answers 7

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We have that

$$3=(3^n)^{1/n}\le (3^n+1)^{1/n}\le (3^n+3^n)^{1/n}=3\cdot2^\frac1n$$

then conclude by squeeze theorem.

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  • $\begingroup$ Sorry if this is a dumb question, but how did you arrive at $\lim_{n \to \infty} (3^n+3^n)^{1/n} $? $\endgroup$
    – user605556
    Aug 17, 2020 at 13:16
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    $\begingroup$ We need something greater than $3^n+1$ to apply squeeze theorem and $3^n+3^n=2\cdot 3^n$ is very helpful since then $(3^n+3^n)^\frac1n =(2\cdot 3^n)^\frac1n=2^\frac1n \cdot 3$. $\endgroup$
    – user
    Aug 17, 2020 at 13:17
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You can use $3=(3^n)^{1/n} \leq(3^n + 1)^{1/n} \leq (3^n+3^n)^{1/n}=2^{1/n}\cdot 3$

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A slightly different way is taking $3^n$ out of $(3^n+1)^{1/n}$, that is $$ (3^n+1)^{1/n}=3(1+3^{-n})^{1/n} $$ Now note that $1\leqslant 1+3^{-n}\leqslant 2$ for every $n\in \mathbb N $, therefore taking limits in the inequality we arrive at $$ 1\leqslant \lim_{n\to\infty}(1+3^{-n})^{1/n}\leqslant \lim_{n\to\infty}2^{1/n}=1 $$ and so $$ \lim_{n\to\infty}(3^n+1)^{1/n}=\lim_{n\to\infty}3(1+3^{-n})^{1/n}=3 $$

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Consider $y = (3^n + 1)^\frac{1}{n}$ now affect logarithm to both sides:$$\ln{y} = \frac{\ln{(3^n + 1)}}{n}$$ obviouslly if $n$ goes to infinity we can omit 1 inside the logarithm then we easily obtain: $\ln{y} = \ln 3$ when $n$ goes to infinity. so the answer is: $$y = 3$$

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    $\begingroup$ Maybe we can enforce the argument by $$\frac{\ln{(3^n + 1)}}{n}= \frac{\ln 3^n+\ln{(1 + 1/3^n)}}{n}=\ln 3+\frac{\ln{(1 + 1/3^n)}}{n}$$ $\endgroup$
    – user
    Aug 17, 2020 at 13:16
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    $\begingroup$ This is exactly what I must add to my answer. Thank you. $\endgroup$ Aug 17, 2020 at 13:20
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With logarithm: rewrite the expression as $$ e^{\frac{1}{n}(\log 3^n + \log (1+\frac{1}{3^n})} $$ The first term is $3$. The second has easy bounds: $$ 0<\log (1+\frac{1}{3^n})<\log 2 $$ and, therefore, $$ 1<e^{\frac{1}{n}\log (1+\frac{1}{3^n})}<e^{\frac{\log 2}{n}} \to_n 1 $$

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  • $\begingroup$ For the second term it suffices to observe that $1/3^n \to 0$. $\endgroup$
    – user
    Aug 17, 2020 at 13:22
  • $\begingroup$ I know. But the OP specifically asked to the squeeze lemma $\endgroup$
    – Alex
    Aug 17, 2020 at 13:25
  • $\begingroup$ Yes I see that but maybe it is a relevant issue for the original expression. Using logarithm, which is a very good alternative way, squeeze theorem seems really not necessary to conclude. $\endgroup$
    – user
    Aug 17, 2020 at 13:28
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$$ \displaystyle\left(3^{n} + 1\right)^{1/n} = 3\left(1 + {1 \over 3^{n}}\right)^{1/n} = 3\left[\left(1 + {1 \over 3^{n}}\right)^{3^{\large n}}\right]^{1/\left(3^{\large n}n\right)} \,\,\,\stackrel{\mathrm{as}\ n\ \to\ \infty}{\to}\,\,\, \bbox[#ffd,10px,border:1px groove navy]{\large 3} $$

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Where $n$ is sufficiently large $3^n$ is much greater that $1$, which can be neglected (we can notice that $100000000000000000000$ and $100000000000000000001$ are "almost" the same).

So $3^n+1 \sim_{n \to \infty} 3^n$ by the fact that $\lim_{n \to \infty} \frac{3^n+1}{3^n}=1$ fastly and the rest can be done easily.

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