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

Question

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?

Answer 1

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 $\int^b_a f(x)\,dx$

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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 $$ \frac{1}{n}\sum^{2n}_{k=1}e^{\frac{k}{n}}=\frac{2}{2n}\sum^{2n}_{k=1}e^{2\tfrac{k}{2n}}\approx 2\int^1_0e^{2u}\,dx$$

Answer 2

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.

Answer 3

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 $$ \int_0^1 e^{2x} \,dx = \frac{1}{2} \lim_{n \to \infty} a_n. $$