find limit of $\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$ with squeeze theorem

Question

I'm trying to prove with squeeze theorem that the limit of the following series equals 1:

$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$

For the left side of the inequality I did:

$$\frac{1+\sqrt{1}+\sqrt[3]{1}+...+\sqrt[n]{1}}{n} < \frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$

For the right side, at first I did the following:

$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n} < \frac{n\sqrt[n]{n}}{n}$$

But then I realized it wasn't true and that the direction of this inequality is the opposite.

Do you have any idea which series with limit 1 is bigger from the original series?

Thanks!

Answer 1

You'll want to know some things about how big $\sqrt[n]{n}$ is. The key facts to prove are:

• For $n$ a positive real number, it's increasing when $n < e$ and decreasing when $n>e$. For integers, $3^{1/3} \approx 1.44$ is the largest value, with $2^{1/2} \approx 1.41$ taking second place.
• As $n \to \infty$, $\sqrt[n]{n} \to 1$. A more precise estimate of $\sqrt[n]{n}$ as $n \to \infty$ is $1 + \frac{\log n}{n}$, but we won't need it.

So we are averaging a few large terms, and many many terms close to $1$. One good way to deal with a situation like that with the squeeze theorem is to separate into two parts: $$ \frac1n \sum_{k=1}^n \sqrt[k]{k} = \frac1n \sum_{k=1}^{\sqrt n}\sqrt[k]k + \frac1n \sum_{k=\sqrt{n}+1}^{n}\sqrt[k]k. $$ What can we say about these two parts?

• In the first sum, we have $\sqrt n$ terms, each of which is at most $3^{1/3}$. So the sum is at most $3^{1/3} \sqrt n$, and we're dividing by $n$. This sum goes to $0$.
• In the second sum, we have nearly $n$ terms, each of which is less than $\sqrt[k]{k}$ for $k = \sqrt n$. So they add up to less than $n \sqrt[k]{k}$. When we divide by $n$, we get $\sqrt[k]{k}$ where $k=\sqrt n$, and this approaches $1$ as $n \to \infty$.

(The specific cutoff of $\sqrt n$ is very flexible: any function $1 \ll f(n) \ll n$ would do.)

Answer 2

As shown in this answer, the Binomial Theorem says that for $n\ge1$, $$ \begin{align} 1\le n^{1/n} &\le1+\sqrt{\frac2n}\tag{1a}\\ &\le1+\frac{2\sqrt2}{\sqrt{n}+\sqrt{n-1}}\tag{1b}\\[3pt] &=1+2\sqrt2\left(\sqrt{n}-\sqrt{n-1}\right)\tag{1c} \end{align} $$ Thus, $$ \frac nn\le\frac1n\sum_{k=1}^nk^{1/k}\le\frac1n\left[n+2\sqrt2\sum_{k=1}^n\left(\sqrt{k}-\sqrt{k-1}\right)\right]\tag2 $$ and, because the sum on the right side of $(2)$ telescopes, we have $$ 1\le\frac1n\sum_{k=1}^nk^{1/k}\le1+\frac{2\sqrt2}{\sqrt{n}}\tag3 $$ to which we can apply the Squeeze Theorem.

Answer 3

This is not a full answer to the question, but many answers are implying that the function $n\mapsto n^{1/n}$ is strictly increasing. This is not the case. To see this:

Let $y=x^{1/x}$. Then $\ln y=\frac 1x \ln x$ so $\frac{y'}{y}=\frac{1}{x^2}(1-\ln x)$. Since $y>0$, this implies that $y$ is increasing on $(0,e)$ and decreasing on $(e,\infty)$.

Hence, do not use upper bound of $n^{1/n}$.

Answer 4

One may combine the following two facts:

1. If $a_n\to a,\,$ then $\,\frac{1}{n}(a_1+\cdots+a_n)\to a$.

2. $\sqrt[n]{n}\to 1$.

Another way to show it is the following: $$ \sqrt[2k]{k}=1+a_k\Longrightarrow \sqrt{k}=(1+a_k)^{k}\ge 1+ka_k \Longrightarrow 0\le a_k<\frac{1}{\sqrt{k}} $$ and hence $$ 1<\sqrt[n]{n}=(1+a_n)^2=1+2a_n+a_n^2<1+\frac{2}{\sqrt{n}}+\frac{1}{n}\le 1+\frac{3}{\sqrt{n}} $$ and thus $$ 1<\frac{1}{n}(1+\sqrt{2}+\cdots+\sqrt[n]{n})<1+\frac{3}{n}\left(1+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}\right) \\ <1+\frac{3}{n}\cdot (2\sqrt{n}+1)\to 1. $$ It remains to show that $$ 1+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}<2\sqrt{n}+1 $$ which can be easily done inductively.

Answer 5

First of all we have, for any $ n\in\mathbb{N}^{*} $, the following : $$ \sqrt[n]{n}=1+\frac{\ln{n}}{n}\int_{0}^{1}{n^{\frac{x}{n}}\,\mathrm{d}x} $$

Since : \begin{aligned}0\leq\int_{0}^{1}{n^{\frac{x}{n}}\,\mathrm{d}x}&\leq n^{\frac{1}{n}}\\ &\leq 2\end{aligned}

We have : \begin{aligned} 1\leq \sqrt[n]{n}\leq 1+\frac{2\ln{n}}{n}&=1+\frac{4\ln{\sqrt{n}}}{n}\\ &\leq 1+\frac{4\sqrt{n}}{n}= 1+\frac{8}{2\sqrt{n}}\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \leq1+\frac{8}{\sqrt{n}+\sqrt{n-1}} \end{aligned}

That remains true for any $ n\in\mathbb{N}^{*} $, which means given $ n\in\mathbb{N}^{*} $, we have : \begin{aligned} 1\leq\frac{1}{n}\sum_{k=1}^{n}{\sqrt[k]{k}}&\leq 1+\frac{8}{n}\sum_{k=1}^{n}{\frac{1}{\sqrt{k}+\sqrt{k-1}}} \\ &\leq 1+\frac{8}{n}\sum_{k=1}^{n}{\left(\sqrt{k}-\sqrt{k-1}\right)}\\ &\leq 1+\frac{8}{\sqrt{n}} \end{aligned}

Thus, using the squeezing theorem, the limit would be $ 1 \cdot$