Limit of sequence: $\lim_{n\to\infty}({2n+1\over 3n})^n$

Question

$\lim_{n\to\infty}({2n+1\over 3n})^n$

For sufficiently large values of $n$:

$0\le ({2n+1\over 3n})^n\le({2\over 3})^n$,

From Squeeze theorem: $\lim_{n\to\infty}({2n+1\over 3n})^n=0$

Is it ok?

I also tried to do this with formula: $\lim_{n\to\infty}(1+{a\over n})^n=e^a$, but with no effect.

Answer 1

No, it is not correct, because the inequality $\left(\frac{2n+1}{3n}\right)^n\leqslant\left(\frac23\right)^n$ never holds. But you can use the fact that$$\lim_{n\to\infty}\left(\frac{2n+1}{3n}\right)^n=\lim_{n\to\infty}\left(\frac23\right)^n\times\lim_{n\to\infty}\left(1+\frac1{2n}\right)^n=0.$$

Answer 2

Note that

$${2n+1\over 3n}\le{2\over 3}$$

is not true, use that eventually

$${2n+1\over 3n}\le \frac45<1 \iff 10n+5<12n \iff 2n>5$$

As a simpler alternative by root test

$$\sqrt[n]{a_n}=\sqrt[n]{\left({2n+1\over 3n}\right)^n}={2n+1\over 3n}\to\frac23<1\implies a_n \to 0$$

Answer 3

Hint: in the original expression take out $\frac{2}{3}$ from your expression inside the brackets and then split.

Answer 4

You can use the root test to get a result quite easily. Since $$\dfrac{2n+1}{3n}\to\dfrac{2}{3}<1$$ by the ratio test $$\bigg(\dfrac{2n+1}{3n}\bigg)^n\to0.$$