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$$ \lim_{n\rightarrow \infty }(3^n+1)^\frac{1}{n} $$

I have been having trouble evaluating this limit using the Squeeze Theorem. Every time I've had to use the squeeze theorem so far there's been a component of the function or sequence which has obviously been bounded by two values, for example $sin(x)$ or $(-1)^\frac {1}{n}$, but I just cant find something like this for this sequence. It's quite obvious the answer is 3, as for when n approaches infinity the 1 will make a negligible contribution to the value of the sequence, so the sequence will effectively just take the n'th root of the n'th power of 3, which is of course 3. The obvious answer is just frustrating me more, as I'm required to use the squeeze theorem to evaluate it, and that I can't seem to figure out. Any help would be very much appreciated!

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    $\begingroup$ $3^n < 3^n + 1 \leqslant 2\cdot 3^n$ $\endgroup$ Aug 6, 2015 at 10:53
  • $\begingroup$ Maybe $3^{(n + 1)/n}$ and $3^{(n - 1)/n}$ as eventual upper & lower bounds? $\endgroup$ Aug 6, 2015 at 10:54

2 Answers 2

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You may write, for $n \geq1$, $$ 3^n < 3^n + 1 <3^n\times 3 $$ giving $$ 3 < \left(3^n + 1 \right)^{1/n}< 3\times 3^{1/n} $$ and you may conclude easily.

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  • $\begingroup$ Thanks for your help, I understand everything except the reasoning behind the $3^n \times 3$, I can see that it works, but I'm unsure as to how you know it would. If you could possibly explain it to me I'd be very grateful. $\endgroup$
    – CoffeeCrow
    Aug 6, 2015 at 11:22
  • $\begingroup$ @CoffeeCrow You have the following inequality:$x+1<3x$ (solve it) which is true as long as $\frac12<x$, applying it with $x=3^n$, $n=1,2,3\cdots$, gives our first inequality (in the answer). Thank you very much! $\endgroup$ Aug 6, 2015 at 13:08
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Also write it out as $$ 3 \cdot 1^n \leq 3(1+\frac{1}{3^n})^{\frac{1}{n}}\leq 3 \cdot (\frac{4}{3})^{\frac{1}{n}} $$ The first and the last sequence converge to $3 \cdot 1$

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