2
$\begingroup$

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!

$\endgroup$
0
7
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Sorry if this is a dumb question, but how did you arrive at $\lim_{n \to \infty} (3^n+3^n)^{1/n} $? $\endgroup$ – jharri Aug 17 '20 at 13:16
  • 2
    $\begingroup$ 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 '20 at 13:17
2
$\begingroup$

You can use $3=(3^n)^{1/n} \leq(3^n + 1)^{1/n} \leq (3^n+3^n)^{1/n}=2^{1/n}\cdot 3$

$\endgroup$
2
$\begingroup$

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, $$ 1<e^{\frac{1}{n}\log (1+\frac{1}{3^n})}<e^{\frac{\log 2}{n}} \to_n 1 $$

$\endgroup$
3
  • $\begingroup$ For the second term it suffices to observe that $1/3^n \to 0$. $\endgroup$ – user Aug 17 '20 at 13:22
  • $\begingroup$ I know. But the OP specifically asked to the squeeze lemma $\endgroup$ – Alex Aug 17 '20 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 '20 at 13:28
2
$\begingroup$

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 $$ \lim_{n\to\infty}(3^n+1)^{1/n}=\lim_{n\to\infty}3(1+3^{-n})^{1/n}=3 $$

$\endgroup$
1
$\begingroup$

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: $$y = 3$$

$\endgroup$
2
  • 2
    $\begingroup$ 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 '20 at 13:16
  • 1
    $\begingroup$ This is exactly what I must add to my answer. Thank you. $\endgroup$ – Amirhossein Dolatkhah Aug 17 '20 at 13:20
1
$\begingroup$

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.

$\endgroup$
0
1
$\begingroup$

$$ \displaystyle\left(3^{n} + 1\right)^{1/n} = 3\left(1 + {1 \over 3^{n}}\right)^{1/n} = 3\left[\left(1 + {1 \over 3^{n}}\right)^{3^{\large n}}\right]^{1/\left(3^{\large n}n\right)} \,\,\,\stackrel{\mathrm{as}\ n\ \to\ \infty}{\to}\,\,\, \bbox[#ffd,10px,border:1px groove navy]{\large 3} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.