# limx→∞(1x∑xi=1(ix)9)\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:

limx→∞(1x∑xi=1(ix)9)\lim_{x\to \infty}\left(\frac{1}{x}\sum_{i=1}^x(\frac{i}{x})^9\right)

I can’t use l’hأ´pitals 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 x9x^9 outside the summation, idk if thats alright?

limx→∞(1×10∑xi=1(i)9)\lim_{x\to \infty}\left(\frac{1}{x^{10}}\sum_{i=1}^x(i)^9\right)

=================

Thanks guys, I just missed it under the stress of the exam I guess
– Marekkk
2 days ago

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.
– grand_chat
2 days ago

The course I’m taking is called precalculus and I should have been able to recognize this as a riemann sum, so why not?
– Marekkk
2 days ago

I regard evaluating ∫10x9dx\int_0^1x^9dx as a calculus exercise. It seems your precalculus course is sneaking in calculus after all 🙂
– grand_chat
2 days ago

=================

6

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This is the definition (in terms of the right-hand Riemann sum) of the definite Riemann integral
∫10x9dx \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 x9x^9, using the power rule x10/10x^{10}/10, evaluate at the two bounds and subtract, giving 1/10.

2

That’s not quite right. This is a possible Riemann sum for the integral.
– qbert
2 days ago

@qbert a Riemann sum doesn’t have a limit as mesh size goes to zero. Only the integral has that.
– ziggurism
2 days ago

I dont’t follow. See en.wikipedia.org/wiki/Riemann_sum
– qbert
2 days ago

3

x is not the index of summation, i is. It is perfectly OK to move the constant out.
– user58697
2 days ago

@user58697 whoops you’re right. Thanks. Lemme edit
– ziggurism
2 days ago

As @ziggurism mentioned, this is most easily evaluated by calculating the definite integral ∫10×9=110.\displaystyle\int_0^1 x^9 = \frac{1}{10}. But if we wanted to do it by factoring a 1×9\dfrac{1}{x^9} out, then we have limx→∞1×10∑1≤i≤xi9=limx→∞1×10120[x2(x+1)2(x2+x−1)(2×4+4×3−x2−3x+3)]=110\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.

en.wikipedia.org/wiki/Faulhaber%27s_formula
– ziggurism
2 days ago

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.

limx→∞(1×10∑xi=1(i)9)\lim_{x\to \infty}\left(\frac{1}{x^{10}}\sum_{i=1}^x(i)^9\right)

Lets look at some easier cases:

∑xi=1(i)=12×2+12x∑xi=1(i)2=13×3+12×2+16x∑xi=1(i)3=14×4+12×2+14x\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:
∑xi=1(i)9=P(x)\sum_{i=1}^x(i)^9 = P(x)
Where P(x)P(x) is a degree 10 polynomial.

How would you find P(x)P(x)?

∑x+1i=1(i)9=∑xi=1(i)9+(x+1)9=P(x+1)\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)9P(x+1) – P(x) = (x+1)^9

P(x)=a10x10+a9x9+a8x8+a7x7⋯+a1x+a0P(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

a10((x+1)10−x10)+a9((x+1)9−x9)⋯+a1((x+1)−x)=(x+1)910a10=19a9=(91)−(102)a108a8=(92)−(92)a9−(103)a10a_{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)\sum_{i=1}^{x+1}(i)^9 = P(x) in a hand-wavey way, we really only need to know a_{10}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}}\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 xx goes to infinity, all the terms but the x^{10}x^{10} terms drop away. And then those cancel, leaving \frac 1{10}\frac 1{10}

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}

\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)(1): Telescoping Sum
(2)(2): Binomial Theorem
(3)(3): change order of summation
(4)(4): pull out the j=1j=1 term and overestimate the remaining sum

We have overestimated the sum in (4)(4) using the Binomial Theorem to get
\sum\limits_{j=0}^m\binom{m}{j}=(1+1)^m=2^m\sum\limits_{j=0}^m\binom{m}{j}=(1+1)^m=2^m and then noting that for j\ge2j\ge2, \sum\limits_{k=1}^nk^{m-j}\le n\cdot n^{m-2}=n^{m-1}\sum\limits_{k=1}^nk^{m-j}\le n\cdot n^{m-2}=n^{m-1}.

Therefore, using Landau Big-O Notation, (4)(4) implies

\sum_{k=1}^nk^{m-1}=\frac1mn^m+O\!\left(n^{m-1}\right)\tag{5}

\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}

\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)(6): pull out the factors of \frac1n\frac1n from the sum
(7)(7): apply (5)(5) with m=10m=10
(8)(8): distribute the \frac1{n^{10}}\frac1{n^{10}}
(9)(9): evaluate the limit

METHODOLOGY 11: Integral Bounds and Application of the Squeeze Theorem

Note that since x^9x^9 is a monotonically increasing, we can bound the sum \frac{1}{N^{10}}\sum_{n=1}^N n^9\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}\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}\lim_{N\to \infty}\frac{1}{N^{10}}\sum_{n=1}^N n^9=\frac1{10}

METHODOLOGY 22: 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}\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)(2) by 10N^{10}10N^{10} and letting N\to \inftyN\to \infty, we obtain the coveted limit.

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

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It’s straightforward evaluated as a Riemann Sum as \texttt{@ziggurism}\texttt{@ziggurism} already shown in a concise fashion.

Another interesting point of view is the Stoltz-Ces\grave{a}\grave{a}ro Theorem
( it doesn’t require to know \underline{explicitly}\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}\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}