Implicit details regarding the proof of $\lim_{n \to \infty}\frac{\sqrt[n]{e^1}+\sqrt[n]{e^2}+\cdots\sqrt[n]{e^n}}{n}$

Question

Problem 9(i) of Spivak's Calculus (Chpt 22) asks for the following limit:

$\displaystyle \lim_{n \to \infty}\frac{\sqrt[n]{e^1}+\sqrt[n]{e^2}+\cdots\sqrt[n]{e^n}}{n}$

The terse solution manual's argument reads as follows:

$\displaystyle \int_0^1 e^x dx = e-1$

Note: My book has not yet covered infinite series

I understand what this argument aims to illustrate, but it seems to me there are several implicit theorems being used here that I do not believe the book has previously addressed (or, at least, not that I am aware of/cannot properly identify). This question is specifically about properly identifying what those theorems are.

Firstly, I see that $\frac{\sqrt[n]{e^1}+\sqrt[n]{e^2}+\cdots\sqrt[n]{e^n}}{n}$ is an upper sum representation of an $n$-subinterval uniform partition $P_n$ on the closed interval $[0,1]$. As such, given that $e^x$ is integrable on this interval, the $\inf$ of the set of all $U(e^x,P)$ exists (where this notation refers to the upper sum of $e^x$ on all partitions defined on $[0,1]$). Next, I know from previous work (If $S$ is set of lower sums for partitions having $n$-equal subintervals, does $\sup S$ equal the $\sup$ of the set of lower sums for all partitions.) that the infimum of the set of all uniform partitions of $[0,1]$ is equal to the infimum of the set of all partitions of $[0,1]$. It seems to me, then, that all I need to do is show that:

$\displaystyle \lim_{n \to \infty}\frac{\sqrt[n]{e^1}+\sqrt[n]{e^2}+\cdots\sqrt[n]{e^n}}{n}=\inf\{U(e^x,P_n):\text{$U$ is an upper sum of $e^x$ and $P_n$ is a uniform partition of $[0,1]$}\}$. If I can prove this, then we are good to go because:

\begin{align}e-1=\int_0^1 e^x dx &=\inf\{U(e^x,P):\text{$U$ is an upper sum of $e^x$ and $P$ is a partition of $[0,1]$}\}\\ &=\inf\{U(e^x,P_n):\text{$U$ is an upper sum of $e^x$ and $P_n$ is a uniform partition of $[0,1]$}\}\\ &=\lim_{n \to \infty}\frac{\sqrt[n]{e^1}+\sqrt[n]{e^2}+\cdots\sqrt[n]{e^n}}{n}\end{align}

So how exactly do I prove that:

$$\displaystyle \lim_{n \to \infty}\frac{\sqrt[n]{e^1}+\sqrt[n]{e^2}+\cdots\sqrt[n]{e^n}}{n}=\inf\{U(e^x,P_n):\text{$U$ is an upper sum of $e^x$ and $P_n$ is a uniform partition of $[0,1]$}\}$$

Through basic partition arguments, I know that:

$U(f,P_{a^0}) \geq U(f,P_{a^1}) \geq U(f,P_{a^2}) \geq \cdots\geq U(f, P_{a^m}) \geq\cdots$ for some $a \in \mathbb N$.

but I also know that just because a subsequence is non-increasing does not, in general, mean that the parent sequence is non-increasing. Calculating the upper sums of uniform partitions for the piece-wise function $f=\begin{cases} 0 \quad &\text{ if $x \in [0,1/3]$}\\1 \quad &\text{ if $x \in (1/3,2/3)$}\\0 \quad &\text{ if $x \in [2/3,1]$}\end{cases}$ is a good example of this.

So although, for example, the subsequence $U(e^x,P_1),U(e^x,P_2),U(e^x,P_4), \cdots, U(e^x, P_{2^m}), \cdots$ is non-increasing, I am unsure if $U(e^x,P_1),U(e^x,P_2),U(e^x,P_3),U(e^x,P_4),U(e^x,P_5),\cdots,U(e^x,P_m),\cdots$ is non-increasing, as well (if I could confirm this, then we are good to go).

While I could show that:

$$\frac{\sqrt[n]{e^1}+\sqrt[n]{e^2}+\cdots\sqrt[n]{e^n}}{n} \geq \frac{\sqrt[n+1]{e^1}+\sqrt[n+1]{e^2}+\cdots\sqrt[n+1]{e^{n+1}}}{n+1} $$

I feel like there must be a more general theorem unrelated to the particular choice of $e^x$ as my function...perhaps if $f$ is continuous then we can say something useful?

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Edit

To flesh out some additional details offered by peek-a-boo, Chapter 13's appendix has the following theorem:

Suppose that $f$ is Darboux integrable on $[a,b]$. Then for every $\varepsilon \gt 0$ there is some $\delta \gt 0$ such that, if $P=\{t_0=a,t_1,\cdots,t_{n-1},t_n=b\}$ is any partition of $[a,b]$ with all lengths $t_i-t_{i-1} \lt \delta$, then: \begin{align} \left| \sum_{i=1}^nf(x_i)(t_i-t_{i-1})-\int_a^b f(x)dx\right| \lt \varepsilon\end{align}

As indicated by peak-a-boo (and others), $\frac{\sqrt[n]{e^1}+\sqrt[n]{e^2}+\cdots\sqrt[n]{e^n}}{n}$ is a Riemann sum of $e^x$ on the closed interval $[0,1]$ where $x_i=\frac{i}{n}$. By the aforementioned theorem, we know that for an arbitrary $\varepsilon$:

$$\left| \frac{\sqrt[n]{e^1}+\sqrt[n]{e^2}+\cdots+\sqrt[n]{e^n}}{n}-\int_0^1 e^xdx\right| \lt \varepsilon$$

where $\frac{1}{n} \lt \delta$.

However, this theorem also tells us that $$\left| \frac{\sqrt[n+j]{e^1}+\sqrt[n+j]{e^2}+\cdots+\sqrt[n+j]{e^{n+j}}}{n+j}-\int_0^1 e^xdx\right| \lt \varepsilon$$

for any $j \in \mathbb N$ because if $1/n \lt \delta$, then $1/(n+j) \lt \delta$.

As such, for any $\varepsilon$, we have found an element ($N=n-1$) such that if $m \gt N: \left| \frac{\sqrt[m]{e^1}+\sqrt[m]{e^2}+\cdots+\sqrt[m]{e^{m}}}{m}-\int_0^1 e^xdx\right| \lt \varepsilon$ , meaning that $\lim_{n \to \infty}\frac{\sqrt[n]{e^1}+\sqrt[n]{e^2}+\cdots+\sqrt[n]{e^n}}{n}=\int_0^1 e^xdx$

Answer 1

A relevant property of $e^x$ is that it is an increasing function in the interval $[0,1]$.

So if you study both the lower as well as the upper sums related to the definite integral $I=\int_0^1e^x\,dx$ (and the partition into $n$ equal length subintervals), you will find that for all $n>1$ we have $$ \frac{e^{0/n}+e^{1/n}+\cdots+e^{(n-1)/n}}n\le I\le \frac{e^{1/n}+e^{2/n}\cdots+e^{(n-1)/n}+e^{n/n}}n. $$ Then look at the difference between the lower and upper bounds as $n\to\infty$.

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Whenever you have sequences of both lower and upper sums with their difference tending to zero, you can deduce that both sequences must converge to the value of the definite integral (and that the function is integrable).

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I am unable to check how much Spivak gets into the finer points of the relation between the Riemann sums as opposed to the upper/lower sums. Judging from the question the latter is used for definitions, so I tailored my answer not to use any result stating that under certain circumstances a sequence of Riemann sums should tend towards the value of the definite integral. After all, such results need extra assumptions about the maximum lengths of the subintervals in the partitions. While easy to verify, those are not needed to settle this.

Answer 2

If $f(x)=e^x$, then $\frac{\sqrt[n]{e^1}+\cdots+\sqrt[n]{e^n}}{n}=\frac{1}{n}\sum_{k=1}^nf\left(\frac{k}{n}\right)=\sum_{k=1}^nf\left(\frac{k}{n}\right)\cdot \frac{1}{n}$. In other words, it is a Riemann sum on $[0,1]$ for the integrable function $f$. So, the limit equals $\int_0^1f=e-1$.

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Edit:

For completeness and context: by this stage, Spivak does introduce the equivalence (in an appendix) of Riemann and Darboux integrability of functions $f:[a,b]\to\Bbb{R}$. To set the notation/terminology,

• the mesh of a partition $P=\{t_0,\dots, t_k\}$ of $[a,b]$ is defined to be $\max\limits_{i\in\{1,\dots, k\}}|t_i-t_{i-1}|$.
• a tagged partition of $[a,b]$ is a pair $(P,\tau)$ where $P=\{t_0,\dots, t_k\}$ is a partition of $[a,b]$ and $\tau$ is a tagging relative to $P$, meaning a collection of points $\{\xi_1,\dots, \xi_k\}$ such that for each $i\in\{1,\dots, k\}$, we have $\xi_i\in [t_{i-1},t_i]$ (i.e a partition, and a choice of one point per subinterval).
• The Riemann sum of $f$ relative to the tagged partition $(P,\tau)$, which we shall denote $\mathcal{R}(f,P,\tau)$ or $\mathcal{R}_f(P,\tau)$ or simply $\mathcal{R}(P,\tau)$ is the sum $\sum_{i=1}^kf(\xi_i)(t_i-t_{i-1})$.

For a Riemann-integrable function $f$, we have that for any sequence of tagged partitions $\{(P_n,\tau_n)\}_{n=1}^{\infty}$, we have \begin{align} \lim_{n\to\infty}\text{mesh}(P_n)=0\implies \lim\limits_{n\to\infty}\mathcal{R}_f(P_n,\tau_n)=\int_a^bf. \end{align} To prove this, let $\epsilon>0$ be given and consider any $\delta>0$ as in the definition of Riemann-integrability. Then, choose $N\in\Bbb{N}$ such that for all $n\geq N$, we have $\text{mesh}(P_n)<\delta$. So, for such $n$, we have $\left|\mathcal{R}_f(P_n,\tau_n)-\int_a^bf\right|<\epsilon$. This proves the claim.

Now, applying this to a uniformly spaced partition shows that $\lim\limits_{n\to\infty}\sum_{i=1}^nf(\xi_i)\cdot\frac{b-a}{n}=\int_a^bf$. Common choices of $\xi_i$’s include (for $i\in\{1,\dots, n\}$)

• $\xi_i=a+(i-1)\frac{b-a}{n}$, i.e the left-Riemann sum
• $\xi_i=a+i\frac{b-a}{n}$, i.e the right-Riemann sum
• the average of the two above, i.e midpoint Riemann sum

In particular, the sum in the question is the right Riemann sum for $f(x)=e^x$ on $[0,1]$, relative to a uniformly-spaced partition. So, with all this theory/preliminaries in mind, this question is indeed a triviality (hence justifying his terse answer).