$\lim_{x\to \infty}\left(\frac{1}{x}\sum_{i=1}^x(\frac{i}{x})^9\right)$ 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?
$\lim_{x\to \infty}\left(\frac{1}{x^{10}}\sum_{i=1}^x(i)^9\right)$
 A: 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.
A: 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.
A: 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 $\frac 1{10}$
A: I thought it might be best to use only ideas from algebra-precalculus.
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
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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}
A: 
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 $p\ge 0$.
