Lower limit of $\sqrt[n]{1^n+2^n+...+n^n}$ with sandwich rule

Question

I was asked to find (by the sandwich rule) the limit of: $$\lim_{n\rightarrow \infty}{\sqrt[n]{1^n+2^n+...+n^n}}$$

The upper limit is quite immidiate: $C_n = \sqrt[n]{n^n+n^n+...+n^n}$.

As for the lower part I'm facing quite an understanding conflict:

I'm looking for $A_n$ for which: $$\forall n\in \mathbb{N}: a_n \leq \sqrt[n]{1^n+2^n+...+n^n}$$ So I wanted to pick $\sqrt[n]{1^n+1^n+...+1^n}$ but I have a dillema: $$\lim_{n\rightarrow \infty}{\sqrt[n]{1^n+1^n+...+1^n}} = 1$$ OR $$\lim_{n\rightarrow \infty}{\sqrt[n]{1^n+1^n+...+1^n}} = \infty$$

1. Since $1$ to any power is $1$ we obviously get: $1^n=1$. Therefor we have $$\lim_{n\rightarrow \infty}{\sqrt[n]{1^n+1^n+...+1^n}} = \lim_{n\rightarrow \infty}{\sqrt[n]{n}} = 1$$

2. Since $n \rightarrow \infty$ we have infinite amounts of $1$'s. Therefore we virtually have $\infty$ under the $\sqrt[n]{}$ sign.

Ultimately I can't seem to determine what's the limit: $\lim_{n\rightarrow \infty}{\sqrt[n]{1^n+1^n+...+1^n}}$.

Is there a ground rule that helps me decide? I'm more prone to decide to go with what I know: $$\lim_{n\rightarrow \infty}{\sqrt[n]{n}} = 1$$ But since I have an inifinite number under the root, Im a bit skeptical. Plus, I already know the answer is $\infty$, but I insist on solving this conflict of mind.

Answer 1

What is $\lim_{n\rightarrow \infty}{\sqrt[n]{1^n+1^n+...+1^n}}$?

For any positive integer $n$, the following is an identity: $$\sqrt[n]{1^n+1^n+...+1^n}=\sqrt[n]{n}$$ The question above is equivalent to ask what is $\lim_{n\rightarrow \infty}{\sqrt[n]{n}}$.

This is a well known limit and it appears in examples in lots of textbooks in real analysis. It is a fact that $\lim_{n\rightarrow \infty}{\sqrt[n]{n}}=1$. The proof in Rudin's Principle of Real Analysis uses the binomial theorem (and the sandwich theorem!). The trick there is writing $x_n=\sqrt[n]{n}-1$ and observe that $$ n=(1+x_n)^n\geq \frac{n(n-1)}{2}x_n^2,\quad x_n\ge 0, $$ which implies $$ 0\le x_n\le \sqrt{\frac{2}{n-1}}. $$

See also this post: Intuitive reason why $\sqrt[n]n\to 1$ as $n\to\infty$?

What goes wrong in the argument for $\lim_{n\rightarrow \infty}{\sqrt[n]{1^n+1^n+...+1^n}}=\infty$?

Since $n \rightarrow \infty$ we have infinite amounts of $1$'s. Therefore we virtually have $\infty$ under the $\sqrt[n]{\phantom{1}}$ sign.

This argument is not even wrong. The symbol $\infty$ is not a real number. Beginners in real analysis should strictly follow definitions regarding any statements containing the symbol $\infty$ or the word "infinite". In the level of "intuition", note that in the expression $$ \sqrt[n]{\underbrace{1^n+\cdots+1^n}_{n \textrm{ terms}}}, $$ not only the radicand $n=\underbrace{1^n+\cdots+1^n}_{n \textrm{ terms}}$ goes larger as $n$ becomes larger, the index $n$ in the radical $\sqrt[n]{\phantom{1}}$ becomes larger too!

What is $\lim_{n\to \infty}{\sqrt[n]{1^n+2^n+...+n^n}}$?

$$ \sqrt[n]{1^n+2^n+...+n^n}\geq \sqrt[n]{n^n}=n $$ which implies that $$ \lim_{n\to \infty}{\sqrt[n]{1^n+2^n+...+n^n}}=\infty. $$

Note that in this case, one cannot apply the sandwich theorem for sequence of real numbers which says the following:

Let ${\displaystyle (a_{n}),(c_{n})}$ be two sequences converging to a real number $\ell$ , and $(b_{n})$ a sequence. If ${\displaystyle \forall n\geqslant N,N\in \mathbb {N} }$ we have $a_{n}\leqslant b_{n}\leqslant c_{n}$, then $(b_{n})$ also converges to $\ell$.

Answer 2

Hints:

• ${\sqrt[n]{1^n+2^n+...+n^n}}\ge \sqrt[n]{n^n} =n$

• $n \to \infty$ as $n \to \infty$

Answer 3

$$\lim_{n\to\infty}\sqrt[n]n$$ is an indeterminate form $\infty^0$.

Taking the logarithm,

$$\log\sqrt[n]n=\frac{\log n}n$$ and the numerator grows much slower than the denominator. By L'Hospital, this limit is also

$$\lim_{n\to\infty}\frac1n.$$

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It is a little funny that you are worried by the fact that the sum $1+1+1\cdots1$ diverges ("virtually $\infty$"), while you don't care about $n$ being exactly that sum !

Answer 4

First of all, it must be clear by now that your sequence's limit is $\;\infty\;$, and if you insist in using the sandwich theorem then you could write

$$\infty\xleftarrow[n\leftarrow\infty]{}\sqrt[n]n\,n=\sqrt{n\cdot n^n}\le\sqrt[n]{1^n+\ldots +n^n}\le\infty$$

In this cases though, in which the sequence is clearly non-negative, it is enough to write the left hand inequality, as the right hand one is trivial.

Answer 5

We can get better inequalities than those presented.

$$ n -1 \; \; \; \; < \; \left( \sum_{j=1}^n \; j^n \right)^{1/n} \; \; \; < n+1 $$

At least two answers point out that the sum is bigger than its final term $n^n,$ taking the $n$th root gives

$$ \color{red}{n \; \; \; \; < \; \left( \sum_{j=1}^n \; j^n \right)^{1/n} \; \; \; < n+1} $$

PROOF:

If we have $f(x) > 0$ and $f'(x) > 0,$ then $$ \int_{a-1}^{b} \; f(x) \; dx \; < \; \sum_{j=a}^b \; f(j) \; < \; \int_{a}^{b+1} \; f(x) \; dx $$

here we are taking $f(x) = x^n,$ then $a=1$ and $b=n.$

$$ \int_{0}^{n} \; x^n \; dx \; < \; \sum_{j=1}^n \; j^n \; < \; \int_{1}^{n+1} \; x^n \; dx $$

$$ n^n \left( \frac{n}{n+1} \right) \; < \; \sum_{j=1}^n \; j^n \; < \; \; (n+1)^n - \frac{1}{n+1} < (n+1)^n $$

$$ n \; \left( \frac{n}{n+1} \right)^{1/n} \; < \; \left( \sum_{j=1}^n \; j^n \right)^{1/n} \; \; \; < n+1 $$

$$ n \; \left( \frac{n}{n+1} \right) \; < \; n \; \left( \frac{n}{n+1} \right)^{1/n} \; < \; \left( \sum_{j=1}^n \; j^n \right)^{1/n} \; \; \; < n+1 $$

$$ n \; \left( \frac{n}{n+1} \right) \; \; < \; \left( \sum_{j=1}^n \; j^n \right)^{1/n} \; \; \; < n+1 $$

NOTE that $$ \frac{n-1}{n} < \frac{n}{n+1} $$

$$n \; \left( \frac{n-1}{n} \right) \; \; \; \; < \; \left( \sum_{j=1}^n \; j^n \right)^{1/n} \; \; \; < n+1 $$

$$n -1 \; \; \; \; < \; \left( \sum_{j=1}^n \; j^n \right)^{1/n} \; \; \; < n+1 $$