How to find the limit of $\sqrt[^n]{n+1}$ as n approaches $\infty$?

Question

How to find the limit of $\sqrt[^n]{n+1}$ as n approaches $\infty$? My teacher hasn't taught L'Hopital's Rule yet, so I am only able to solve this using some standard limits.

The similar standard limit I can think of is $\lim_{n\to\infty} (1+\frac{a}{n})^n = e^a$.

What I have done so far:

$\lim_{n\to\infty} \sqrt[^n]{n+1}$

$ = \lim_{n\to\infty} (n+1)^{\frac{1}{n}}$

Let $ m = \frac{1}{n} \therefore n = \frac{1}{m}$. Since $n\rightarrow\infty$, $m\rightarrow0$.

$\lim_{m\to0} (1+\frac{1}{m})^m$

But I'm stuck in this step because I don't know how to work out the answer. If I just substitute $0$ in m, $1+\frac{1}{m}$ is undefined. If I rewrite this as $e^{log((1+\frac{1}{m})^m)}$, it's also undefined. I know the limit is $1$ by looking at the graph. I'm wondering how I'm able to solve it in a more rigorous way.

Answer 1

From copper.hat's hint: \begin{equation} \begin{split} \lim_{n\to\infty} \sqrt[^n]{n+1}&=\lim_{n\to\infty} (n+1)^{\frac{1}{n}}\\ &= \lim_{n\to\infty} e^{\log((n+1)^{\frac{1}{n}})}\\ &= \lim_{n\to\infty} e^{({\frac{1}{n}\log(n+1)})}\\ \end{split} \end{equation}

Let $t = \log(n+1)$. We have that

\begin{equation} \begin{split} &e^t = e^{\log(n+1)}=n+1\\ &n = e^t - 1\\ \end{split} \end{equation}

Therefore, our limit equals

$$\displaystyle \lim_{t\to\infty} e^{\frac{t}{e^t-1}}= \large{e}^{\displaystyle \small{\lim_{t\to\infty} \frac{t}{e^t-1}}} = e^0 = 1$$

The limit of the exponent is $0$ because the denominator $e^t$ grows much faster than the numerator $t$.

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Another solution:

\begin{equation} \begin{split} \lim_{n\to\infty} \sqrt[^n]{n+1}&=\lim_{n\to\infty} (n+1)^{\frac{1}{n}}\\ &= \lim_{n\to\infty} e^{\log((n+1)^{\frac{1}{n}})}\\ &= \lim_{n\to\infty} e^{({\frac{log(n+1)}{n}})}\ \end{split} \end{equation}

The limit of the exponent is $0$ because the denominator $n$ grows much faster than the numerator $log(n+1)$.

Answer 2

There is one elementary approach, using only the bionomial theorem.

We know that for $n\in\mathbb{N}$, $n+1>1$ and therefore $\sqrt[n]{n+1}>1$. Suppose that it does not converge to $1$, that is, $\exists\varepsilon>0$, $\exists\{n_j\}_{j\in\mathbb{N}}$ s.t $\sqrt[n_j]{n_j+1} > 1+\varepsilon$ (it cannot be smaller than $1-\varepsilon$ by the above inequality.) Then we have $$n_j+1>(1+\varepsilon)^{n_j} \ge 1+n_j\varepsilon+\frac{n_j(n_j-1)}{2}\varepsilon^2$$ (for $n_j\ge 2$.) As $j\to\infty$, $n_j\to\infty$, and this inequality eventually fails to hold. This is a contradiction, and we know that the sequence converges to $1$.

Answer 3

Let $x_n$ be the sequence given by

$$\begin{align} x_n&=(n+1)^{1/n}\tag1 \end{align}$$

From $(1)$ we find that $n=x_n^n-1$ and we must have $x_n\ge 1$. Writing $x_n=1+y_n$ with $y_n\ge 0$, applying the binomial theorem, and rearranging reveals for $n>1$

$$0\le y_n\le \sqrt{\frac2{n-1}}\tag 2$$

Applying the squeeze theorem to $(2)$ shows that $\lim_{n\to\infty}y_n=0$ and hence

$$\lim_{n\to\infty}x_n=1$$