Can't figure out how to work out the correct partition to translate an upper sum into the correct integral for evaluation. Spivak Ch.22 - 9) ii This question is directly related to a previous question I asked:
Question from *Spivak Calculus*, Ch.22 - #9 i) - relationship between sequences and integration.
My problem is I can't seem to work out how to get the correct partition for the following upper sum which I'm equating to a sequence.
The original expression we are asked to evaluate is:
$$\lim_{n \to \infty}\frac{\sqrt[n]{e} + \sqrt[n]{e^{2}} + \dots + \sqrt[n]{e^{2n}}}{n}$$
EDIT: This is the question directly from the textbook (here I am asking about part (ii):

A formula that can be deduced for this is:
$$a_{n} = \frac{1}{n}\sum_{i = 1}^{n}(e^{\frac{i}{n}})^{2}$$
The problem I'm having is figuring out the correct way to express the partition. For the original question I found the formula to be:
$$a_{n} = \frac{\sum_{i = 1}^{n}e^{\frac{i}{n}}}{n}$$
From which we can determine a partition to be of length $\frac{1}{n}$ over the interval $[0,1]$ (I think my original issue is I'm not even sure how this interval was determined to be the "right" one). But with this idea we arrive at:
$$\int_{x=0}^1 e^x \, dx = \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n e^{i/n}.$$
I know for this question the interval is supposed to be $[0,2]$, but I haven't been able to formally work it out. From the expression I have above what I was expecting to arrive at was something of the form:
$$a_{n} = \frac{2}{n}\sum_{i = 1}^{n}(e^{\frac{i}{n}})^{2}$$
From which the interval would could be instantly seen. But that didn't occur for me. What is it that I'm missing in being able to work out the correct intervals?
 A: Partition the interval $[0,1]$ into $2n$ evenly spaced subintervals, so the partition is $\{0, \frac{1}{2n}, \frac{2}{2n}, \dots, \frac{2n-1}{2n}, 1 \}$.  Let $f(x) = e^{2x}$.  Then the Riemann sum for $f$ over $[0,1]$, using the right endpoints, is
$$
\sum_{i=1}^{2n} \frac{1}{2n} f(i/2n) = \frac{1}{2n} \sum_{i=1}^{2n} e^{i/n} = \frac{1}{2} a_n.
$$
(where $a_n$ is the formula for the $n$th term of the sequence in part (ii) of the question you posted).  Taking the limit as $n \to \infty$ you get
$$
\int_0^1 e^{2x} \,dx = \frac{1}{2} \lim_{n \to \infty} a_n.
$$
A: We define
$$
b_n = \frac1n \sum_{i=1}^{2n} (e^{i/n}).
$$
We want to calculate $\lim_{n\to\infty}$ using the method you proposed,
using the interval $[0,2]$
We have
$$
b_n = \frac2{2n} \sum_{i=1}^{2n} e^{(2i)/(2n)}.
$$
If we define $c_m$ via
$$
c_m = \frac1{m} \sum_{i=1}^{m} 2e^{(2i)/m}
$$
then one can see that $b_n=c_{2n}$.
Note that $b_n$ has the same limit as $c_m$ (if $c_m$ converges),
so we only need to calculate $\lim c_m$.
Then you can calculate $\lim c_{m}$ using the usual method:
We use the function $f(x)=2(e^{x})$ on the interval
$[0,2]$, and partition this interval into $n$ equal parts.
Then we have
$
c_n = \frac1n \sum_{i=1}^n f((2i)/n),
$
which corrresponds to the integral $\int_0^2 f(x)$.
What interval is the right one?:
There is not always a right interval, and if one wishes, one can also use other intervals.
I used $[0,2]$ since it was mentioned in the question.
A good way to find the interval is to first guess the function
(in this case $f(x)=e^x$, but other functions would be also possible).
Then the interval boundaries $s,t$ should be chosen such that
$f(s)$ is close to the first summand (here, $e^{1/n} \sim e^0$ for large $n$),
and $f(t)$ is close to the last summand (here, $e^{{2n}/n} \sim e^2$ for all $n$).
Then one would choose the interval $[s,t]$ and continue working from there.
A: Recall that for a partition $x_k:=a+\frac{b-a}{N}k$, $k=0,\ldots,M$ , of an interval $[a,b]$ and a function $f$,
$$
\frac{b-a}{N}\sum^N_{k=1}f(x_k)=h\sum^N_{k=1}f\big(a+kh), \qquad h=\frac{b-a}{N}
$$
is a Riemann approximation to $\int^b_a f(x)\,dx$

For (ii), $\frac{1}{n}\sum^{2n}_{j=1}e^{i/n}$ can be seen as a Riemann sum for the integral $\int^2_0e^{x}\,dx$.

*

*Consider  the partition $x_k=\frac{k}{n}$, $k=0,\ldots,2n$, of the interval $I=[0,2]$ (divide $I$ in $2n$ pieces of the same length.

*The Riemann sum one gets by taking the right-hand endpoints if the subintervals $[x_{k-1},x_k]$ ($k=1,\ldots,n$) generated by this partition is
$$\frac{2}{2n}\sum^{2n}_{k=1}e^{\tfrac{2}{2n}k}\approx\int^2_0 e^x\,dx=2\int^1_0 e^{2u}\,du$$
via the change of variables $x=2u$.

*Not surprisingly, the  sum $\frac{1}{n}\sum^{2n}_{k=1}e^{\frac{k}{n}}$ can also bee seen as a Riemann sum for the integral  $2\int^1_0 e^{2x}\,dx$. Use the partition $t_k=\frac{k}{2n}$, $k=0,\ldots,2n$ of the interval $[0,1]$.  Then
$$
\frac{1}{n}\sum^{2n}_{k=1}e^{\frac{k}{n}}=\frac{2}{2n}\sum^{2n}_{k=1}e^{2\tfrac{k}{2n}}\approx 2\int^1_0e^{2u}\,dx$$
