# Evaluate the limit $\lim_{n \to \infty} \left(3^n+1\right)^{\frac1n}$

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!

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.

• Sorry if this is a dumb question, but how did you arrive at $\lim_{n \to \infty} (3^n+3^n)^{1/n}$? – jharri Aug 17 '20 at 13:16
• 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 $(3^n+3^n)^\frac1n =(2\cdot 3^n)^\frac1n=2^\frac1n \cdot 3$. – user Aug 17 '20 at 13:17

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

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

• For the second term it suffices to observe that $1/3^n \to 0$. – user Aug 17 '20 at 13:22
• I know. But the OP specifically asked to the squeeze lemma – Alex Aug 17 '20 at 13:25
• 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. – user Aug 17 '20 at 13:28

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

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: $$y = 3$$

• Maybe we can enforce the argument by $$\frac{\ln{(3^n + 1)}}{n}= \frac{\ln 3^n+\ln{(1 + 1/3^n)}}{n}=\ln 3+\frac{\ln{(1 + 1/3^n)}}{n}$$ – user Aug 17 '20 at 13:16
• This is exactly what I must add to my answer. Thank you. – Amirhossein Dolatkhah Aug 17 '20 at 13:20

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.

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