$ \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 $a=1$

Answer 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?

Answer 2

1. 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 $0 \le a < 1$.

2. 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 $a$.

3. 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) $.

Answer 3

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 $a \cdot 1 = a$.

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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.

Answer 4

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.