Does the nth root of $n \cdot 2^{3n}+3^{2n}$ converge as $n \to \infty$?

Question

Consider the sequence

$a_n = \sqrt[n]{n \cdot 2^{3n}+3^{2n}}$, $n\in\mathbb{N}$.

With a string of inequalities, one can show that $a_n$ is bounded and the graph of the function $ f(x) = \sqrt[x]{x \cdot 2^{3x}+3^{2x}}$ suggests that $f$ is monotone, but how could one prove convergence and a calculate the limit of $a_n$?

I would be grateful for any help!

Hint. $a_n = 9\sqrt[n]{n \cdot (8/9)^n+1}$ – Robert Z Feb 28 '22 at 11:29 — Robert Z, Feb 28 '22 at 11:29

cos_dm_math21 · Answer 1 · 2022-02-28T12:05:12.863

1

Note that $$a_n = \sqrt[n]{n\cdot 8^n+9^n} = 9\cdot \sqrt[n]{n\cdot \left(\frac{8}{9}\right)^n+1}$$ Denote by $ \displaystyle b_n = n\cdot \left(\frac{8}{9}\right)^n$. Since $$\frac{b_{n+1}}{b_n} = \frac{n+1}{n}\cdot \frac{8}{9} \longrightarrow \frac{8}{9}<1$$ it follows that $b_n \longrightarrow0$, and thus $a_n \longrightarrow 9$.

Addendum: Since $1\le b_n+1$ it follows that $1\le \sqrt[n]{b_n+1}\le b_n+1$. Thus, by the squeezing principle, $\sqrt[n]{b_n+1}$ tends to $1$, as I claimed.

edited Feb 28 '22 at 12:05

answered Feb 28 '22 at 11:35

cos_dm_math21

1,895

$\lim 1/2^n=0$ but $\lim \sqrt[n]{1/2^n}=1/2$ – markvs Feb 28 '22 at 11:39
How about lim $\sqrt[n]{1/2^n+1}$? – cos_dm_math21 Feb 28 '22 at 11:42
It is $1$. What about it? – markvs Feb 28 '22 at 11:48
I don't get the idea of your first comment. I thought you are trying to point out a mistake in my argument. – cos_dm_math21 Feb 28 '22 at 11:50
Your argument is wrong because from $a_n\to a$ does not follow that $\sqrt[n]{a_n}\to 1$. You claim without a proof that since $b_n+1\to 1$ then $\sqrt[n]{b_n+1}=1$. – markvs Feb 28 '22 at 11:53
Please read my proof carefully. What I use in the proof is the following: If a sequence $x_n$ tends to $0$, then $\sqrt[n]{x_n+1}$ tends to $1$. – cos_dm_math21 Feb 28 '22 at 11:56
As I said, you did not prove this fact. It is wrong if we replace the first $1$ by $0$. – markvs Feb 28 '22 at 11:58

markvs · Accepted Answer · 2022-02-28T11:49:19.427

1

Rewrite as $\lim 8\sqrt[n]{1+(9/8)^n}\ge \lim 8\sqrt[n]{(9/8)^n}=9$ and $$\lim 8\sqrt[n]{n+(9/8)^n}\le \lim 8\sqrt[n]{2(9/8)^n}=9.$$ So the limit is $9$.

edited Feb 28 '22 at 11:49

answered Feb 28 '22 at 11:43

markvs

19,653

It is $\sqrt[n]{n+(9/8)^n}$ instead of $\sqrt[n]{1+(9/8)^n}$. – cos_dm_math21 Feb 28 '22 at 11:48
Yes, it does not matter because $(9/8)^n\le n+(9/8)^n\le 2(9/8)^n$ for large enough $n$. I have fixed it. – markvs Feb 28 '22 at 11:51

Does the nth root of $n \cdot 2^{3n}+3^{2n}$ converge as $n \to \infty$?

2 Answers2