3
$\begingroup$

For precalculus I had an exam today and we had to solve the following question:

$\lim_{x\to \infty}\left(\frac{1}{x}\sum_{i=1}^x(\frac{i}{x})^9\right)$

I can't use l'hôpitals rule to solve this. From calculating with a large value for x I know the value must be around 1, I just need to prove it.

I thought it could be solved by moving the $x^9$ outside the summation, idk if thats alright?

$\lim_{x\to \infty}\left(\frac{1}{x^{10}}\sum_{i=1}^x(i)^9\right)$

$\endgroup$
4
  • $\begingroup$ Thanks guys, I just missed it under the stress of the exam I guess $\endgroup$
    – Marekkk
    Commented Oct 21, 2016 at 23:20
  • $\begingroup$ Is this a precalculus question? If you don't know the sum of ninth powers formula, and are not expected to derive it, then this problem is mighty hard without calculus. $\endgroup$
    – grand_chat
    Commented Oct 21, 2016 at 23:29
  • $\begingroup$ The course I'm taking is called precalculus and I should have been able to recognize this as a riemann sum, so why not? $\endgroup$
    – Marekkk
    Commented Oct 21, 2016 at 23:33
  • $\begingroup$ I regard evaluating $\int_0^1x^9dx$ as a calculus exercise. It seems your precalculus course is sneaking in calculus after all :) $\endgroup$
    – grand_chat
    Commented Oct 21, 2016 at 23:51

6 Answers 6

3
$\begingroup$

This is the definition (in terms of the right-hand Riemann sum) of the definite Riemann integral $$ \int_0^1 x^9\,dx$$

Did you learn how to evaluate such integrals?

Using the fundamental theorem of calculus, you find an antiderivative of $x^9$, using the power rule $x^{10}/10$, evaluate at the two bounds and subtract, giving 1/10.

$\endgroup$
9
  • 2
    $\begingroup$ That's not quite right. This is a possible Riemann sum for the integral. $\endgroup$ Commented Oct 21, 2016 at 23:07
  • $\begingroup$ @qbert a Riemann sum doesn't have a limit as mesh size goes to zero. Only the integral has that. $\endgroup$
    – ziggurism
    Commented Oct 21, 2016 at 23:09
  • $\begingroup$ I dont't follow. See en.wikipedia.org/wiki/Riemann_sum $\endgroup$ Commented Oct 21, 2016 at 23:11
  • 3
    $\begingroup$ x is not the index of summation, i is. It is perfectly OK to move the constant out. $\endgroup$
    – user58697
    Commented Oct 21, 2016 at 23:11
  • $\begingroup$ @user58697 whoops you're right. Thanks. Lemme edit $\endgroup$
    – ziggurism
    Commented Oct 21, 2016 at 23:12
2
$\begingroup$

As @ziggurism mentioned, this is most easily evaluated by calculating the definite integral $\displaystyle\int_0^1 x^9 = \frac{1}{10}.$ But if we wanted to do it by factoring a $\dfrac{1}{x^9}$ out, then we have $\displaystyle \lim_{x\to\infty} \frac {1}{x^{10}}\sum_{1\leq i\leq x} i^9 = \lim_{x\to\infty} \frac{1}{x^{10}}\frac{1}{20} \left[x^2 (x+1)^2 (x^2+x-1) (2 x^4+4 x^3-x^2-3 x+3)\right] = \frac{1}{10}$ This approach however requires you know the sum of 9th powers formula.

$\endgroup$
1
1
$\begingroup$

They are teaching you calculus in your pre-calculus class. I don't think they are actually doing you a service. Better to do this problem at a lower degree of difficulty then learn the fundamental theorem of calculus and apply it to the more difficult cases.

Everything looks great so far.

$\lim_{x\to \infty}\left(\frac{1}{x^{10}}\sum_{i=1}^x(i)^9\right)$

Lets look at some easier cases:

$\sum_{i=1}^x(i) = \frac 12 x^2 + \frac 12 x\\ \sum_{i=1}^x(i)^2 = \frac 13 x^3 + \frac 12x^2 + \frac 16 x\\ \sum_{i=1}^x(i)^3 = \frac 14 x^4 + \frac 12 x^2 + \frac 14 x$

I am going to suggest: $\sum_{i=1}^x(i)^9 = P(x)$ Where $P(x)$ is a degree 10 polynomial.

How would you find $P(x)$?

$\sum_{i=1}^{x+1}(i)^9 = \sum_{i=1}^{x}(i)^9 + (x+1)^9 = P(x+1)$

$P(x+1) - P(x) = (x+1)^9$

$P(x) = a_{10} x^{10} + a_9 x^9 + a_8 x^8 + a_7 x^7 \cdots + a_1 x + a_0$ is a generic degree 10 polynomial

$a_{10} ((x+1)^{10} - x^{10}) + a_9 ((x+1)^9 - x^9)\cdots +a_1((x+1) - x) = (x+1)^9\\ 10 a_{10} = 1\\ 9a_9 = {9\choose 1} - {10\choose 2}a_{10}\\ 8a_8 = {9\choose 2} - {9\choose 2}a_9 - {10\choose 3}a_{10}$

etc.

And in this way you can find the full polynomial. But, it is not entirely necessary. If we can accept that $\sum_{i=1}^{x+1}(i)^9 = P(x)$ in a hand-wavey way, we really only need to know $a_{10}$

$\lim_{x\to \infty}\left(\frac{1}{x^{10}}\sum_{i=1}^x(i)^9\right) = \frac {\frac 1{10}x^{10} + a_9 x^9 + a_8 x^8\cdots + a_1 x}{x^{10}}$

As $x$ goes to infinity, all the terms but the $x^{10}$ terms drop away. And then those cancel, leaving $\frac 1{10}$

$\endgroup$
1
$\begingroup$

I thought it might be best to use only ideas from .

Note that $$ \begin{align} n^m &=\sum_{k=1}^nk^m-\sum_{k=1}^n(k-1)^m\tag{1}\\ &=\sum_{k=1}^n\sum_{j=1}^m(-1)^{j-1}\binom{m}{j}k^{m-j}\tag{2}\\ &=\sum_{j=1}^m(-1)^{j-1}\binom{m}{j}\sum_{k=1}^nk^{m-j}\tag{3}\\ &=m\sum_{k=1}^nk^{m-1}+\underbrace{\sum_{j=2}^m(-1)^{j-1}\binom{m}{j}\sum_{k=1}^nk^{m-j}}_{\le2^mn^{m-1}=O\left(n^{m-1}\right)}\tag{4} \end{align} $$ Explanation:
$(1)$: Telescoping Sum
$(2)$: Binomial Theorem
$(3)$: change order of summation
$(4)$: pull out the $j=1$ term and overestimate the remaining sum

We have overestimated the sum in $(4)$ using the Binomial Theorem to get $\sum\limits_{j=0}^m\binom{m}{j}=(1+1)^m=2^m$ and then noting that for $j\ge2$, $\sum\limits_{k=1}^nk^{m-j}\le n\cdot n^{m-2}=n^{m-1}$.

Therefore, using Landau Big-O Notation, $(4)$ implies $$ \sum_{k=1}^nk^{m-1}=\frac1mn^m+O\!\left(n^{m-1}\right)\tag{5} $$ Thus, $$ \begin{align} \lim_{n\to \infty}\left(\frac1n\sum_{k=1}^n\left(\frac kn\right)^9\right) &=\lim_{n\to \infty}\left(\frac1{n^{10}}\sum_{k=1}^nk^9\right)\tag{6}\\ &=\lim_{n\to \infty}\left(\frac1{n^{10}}\left[\frac1{10}n^{10}+O\!\left(n^9\right)\right]\right)\tag{7}\\[4pt] &=\lim_{n\to \infty}\left(\frac1{10}+O\!\left(\frac1n\right)\right)\tag{8}\\[5pt] &=\frac1{10}\tag{9} \end{align} $$ Explanation:
$(6)$: pull out the factors of $\frac1n$ from the sum
$(7)$: apply $(5)$ with $m=10$
$(8)$: distribute the $\frac1{n^{10}}$
$(9)$: evaluate the limit

$\endgroup$
1
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ It's straightforward evaluated as a Riemann Sum as $\texttt{@ziggurism}$ already shown in a concise fashion.


Another interesting point of view is the Stoltz-Ces$\grave{a}$ro Theorem ( it doesn't require to know $\underline{explicitly}$ the sum ): \begin{align} \lim_{x\to \infty}\bracks{{1 \over x}\sum_{i\ =\ 1}^{x}\pars{i \over x}^{9}} & = \lim_{x\to \infty}\pars{{1 \over x^{10}}\sum_{i\ =\ 1}^{x}i^{9}} = \lim_{x\to \infty}{\sum_{i\ =\ 1}^{x + 1}i^{9} - \sum_{i\ =\ 1}^{x}i^{9} \over \pars{x + 1}^{10} - x^{10}} \\[5mm] & = \lim_{x\to \infty}{\pars{x + 1}^{9} \over \sum_{k = 0}^{8}{10 \choose k}x^{k} + {10 \choose 9}x^{9}} = \lim_{x\to \infty}{\pars{1 + 1/x}^{9} \over \sum_{k = 0}^{8}{10 \choose k}\pars{1/x}^{9 - k} + {10 \choose 9}} = {1 \over {10 \choose 9}} \\[5mm] & = \bbox[#ffe,10px,border:1px dotted navy]{\ds{1 \over 10}} \end{align}

$\endgroup$
0
$\begingroup$

METHODOLOGY $1$: Integral Bounds and Application of the Squeeze Theorem

Note that since $x^9$ is a monotonically increasing, we can bound the sum $\frac{1}{N^{10}}\sum_{n=1}^N n^9$ by

$$\frac{1}{10}=\frac{1}{N^{10}}\int_0^N x^9\,dx \le \frac{1}{N^{10}}\sum_{n=1}^N n^9\le \frac{1}{N^{10}}\int_1^{N+1}x^9\,dx=\frac{(1+1/N)^{10}-1/N^{10}}{10}$$

whereupon applying the squeeze theorem we obtain the coveted limit

$$\lim_{N\to \infty}\frac{1}{N^{10}}\sum_{n=1}^N n^9=\frac1{10}$$


METHODOLOGY $2$: Summation by Parts

If one does not wish to pursue the evaluation using integration principles, the we can instead procees using Summation by Parts.

Proceeding, we have

$$\begin{align} \sum_{n=1}^N n^9&=(N+1)^{10}-1-\sum_{n=1}^N (n+1)\left( (n+1)^9-n^9 \right)\\\\ &=(N+1)^{10}-1-\sum_{n=1}^N (n+1)(9n^8+O(n^7)) \tag 1\\\\ 10\sum_{n=1}^N n^9&=(N+1)^{10}-1+O(N^9)\tag 2 \end{align}$$

Dividing both sides of $(2)$ by $10N^{10}$ and letting $N\to \infty$, we obtain the coveted limit.

Note that in arriving at $(1)$ we simply applied the binomial theorem, while in deducing $(2)$ we made use of the fact that $\sum_{n=1}^N n^p\le N^{p+1}$ for $p\ge 0$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .