$ \lim_{n \to \infty} \sqrt[n]{a^n+1}$ with $a \ge 0$

Question

I'm a bit rusty with limits $$ \lim_{n \to \infty} \sqrt[n]{a^n+1}$$ with $a \ge 0$.

The solution in my book is $max \left \{ 0,1 \right \}$ but my final results are:

1) $+\infty$ if $0<a<1$

2) $1$ if $a>1$

3) $+\infty$ if $a=1$

Note that $\max{0,1} = 1$ -- presumably you meant something else? — Aaron Montgomery, Nov 27 '17 at 16:26
Yes because $$ \sqrt[n]{a^n+1}=(a^n+1)^{\frac{1}{n}}$$ and $a^b=e^{b\log(a)}$ (with conditions). If $0<a<1$ then $a^n\to 0$ use $\log(1+h)$ with $h$ small, if $a>1$ use $(a^n+1)=a^n(1+\frac{1}{a^n})$. The case $a=1$ is straightforward. — Duchamp Gérard H. E., Nov 27 '17 at 16:32
if $|a|<1$ then $\frac1n\ln(1+a^n)=\sum_{k\ge 1}(-1)^{k+1}\frac{a^{nk}}{nk}$ — Masacroso, Nov 27 '17 at 16:34
If $a\ge1$ lim = $a$ and if $a\le 1$ lim =$1$ so answer is $max{a, 1}$ all these limits are straight forward. — Stephen Meskin, Nov 27 '17 at 16:37
@AaronMontgomery, I'm guessing the $0$ in $\max{0,1}$ is supposed to be an $a$. — Barry Cipra, Nov 27 '17 at 16:42

score 4 · Answer 1 · answered Nov 27 '17 at 16:40

The correct result is $\max\{a, 1\}$, so I suppose it's a typo or a miscopied expression. To see this, note that

If $a \le 1$, then $1 \le a^n + 1 \le 2$ for all $n$. Taking $n$-th roots and a limit gives limit $1$.
If $a > 1$, then $$\left(a^n + 1\right)^{1/n} = a \left(1 + \frac{1}{a^n}\right)^{1/n}$$ Now apply the previous case with $1/a$ to see why this tends to $a \cdot 1 = a$.

The key idea here is that if $a < 1$, the term $a^n$ is negligible once $n$ is large. If $a > 1$, the term $1$ is negligible once $n$ is large.

lhf · Answer 2 · 2017-11-27T16:40:49.467

3

By Bernoulli's inequality$, 1 \le (1+a^n)^{1/n}\leq 1+\frac{a^n}{n} \to 1$ and so the limit is $1$ if $0 \le a < 1$.
If $a>1$, then $a \le \sqrt[n]{a^n+1} \le \sqrt[n]{a^n+a^n} = a \sqrt[n]{2} \to a$ and so the limit is $a$.
If $a=1$, then $\sqrt[n]{a^n+1}=\sqrt[n]{2} \to 1$.

You don't really need case 3 because the other two cases work for $a=1$.

Therefore, $\displaystyle \lim_{n \to \infty} \sqrt[n]{a^n+1} = \max(a,1) $.

edited Nov 27 '17 at 16:40

answered Nov 27 '17 at 16:35

lhf

and for $a>1$ $lim_{n->\infty}(a^n+1)^{\frac{1}{n}}=a$. – chak Nov 27 '17 at 16:40
In 2: $\sqrt[n]{a^n+1}=\sqrt[n]{a^n}\sqrt[n]{(1+a^{-n})}\to a$, no need to inequality – ℋolo Nov 27 '17 at 16:46

score 1 · Accepted Answer · answered Nov 27 '17 at 17:31

1

1) $0<a<1:$

$1 \lt (a^n +1)^{1/n} \lt (2)^{1/n} .$

$1\le \lim_{n \rightarrow \infty} (a^n+1)^{1/n} \le $

$\lim_{ n\rightarrow \infty }2^{1/n} =1.$

2) $a>1:$

$a \lt (a^n +1)^{1/n} \lt (2a^n)^{1/n}.$

$a \le \lim_{ n \rightarrow \infty}(a^n+1)^{1/n} \le $

$\lim_{n \rightarrow \infty}(2)^{1/n}a = a.$

3)$a=1. $

Can you do it?

answered Nov 27 '17 at 17:31

Peter Szilas

I'm not used to the bernoully 's inequality ...lim_n a^n=0 , does (a^n+1)^{1/n} should be 1^{1/n} – Anne Nov 27 '17 at 17:54
Anne. y= x^1/n is strictly increasing: means: x <y, then x^1/n < y^1/n. First line:: 1<a^n+1<(1+1) since a^n <1; now take the n-th root. The case a>1 : a^n <a^n +1 < 2a^n, since a^n>1. Now take the root. Does this make sense? – Peter Szilas Nov 27 '17 at 18:56

score 1 · Answer 4 · answered Nov 27 '17 at 17:47

I agree with $\sqrt[n]{a^n+1}=e^{log\sqrt[n]{a^n+1} }=e^{\frac{log(a^n+1)}{n} }$

If $a>1$ $\frac{log(a^n+1)}{n}<\frac{log(a^n)}{n}=\frac{n*log(a)}{n}=log(a)$ and then $lim_n \sqrt[n]{a^n+1}=e^{log {a}}=a$

If $0<a<1$ $\frac{log(a^n+1)}{n}<\frac{log(a^n)}{n}=\frac{n*log(a)}{n}=log(a)$ and then $lim_n \sqrt[n]{a^n+1}=e^{log {a}}=a$ (but $a<1$!!!)

If $a=1$ , $\sqrt[n]{a^n+1}=\sqrt[n]{1^n+1} $and then $lim_n \sqrt[n]{1^n+1}=lim_n \sqrt[n]{2}=1$

If $a=0$ , $\sqrt[n]{a^n+1}=\sqrt[n]{1} $ and then $lim_n \sqrt[n]{1}=1$

I'm not sure about the case of $0<a<1$ because a <1 and the result is not the maximum between a and 1.

4 Answers4