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I am working on a problem and I feel like I may not completely understand that concept which I think is real important that I do.

The question is in regard to showing the validity of the following inequality
$\int_0^{n}x^{q}dx$ $\le 1+2^{q}+..+n^{q} \le \int_0^{n+1}x^{q}dx$ for any positive integer n and any real q $\ge 0 $

One answer, similar to one suggested by the user Tryss on this site is as follows,

$$\int_0^n x^q dx = \sum_{k=1}^{n} \int_{k-1}^k x^q dx \leq \sum_{k=1}^{n} \int_{k-1}^k k^q dx $$

Then

$$\int_0^n x^q dx \leq \sum_{k=1}^{n} k^q = 1+ 2^q + \cdots n^q $$

And we have,

$$\int_0^{n+1} x^q dx = \sum_{k=0}^{n} \int_{k}^{k+1} x^q dx \geq \sum_{k=0}^{n} \int_{k}^{k+1} k^q dx $$

i,e then,

$$\int_0^{n+1} x^q dx \geq \sum_{k=0}^{n} k^q = 1+ 2^q + \cdots n^q $$

The next part of the question that I am mostly stumped on is to find $\lim_{n\to \infty}a_n$ where $a_n= 1/{n^{q+1}}+2^{q}/n^{q+1}+…+n^{q}/n^{q+1}$. I'm sure I need to use the result of the first part of the problem but I am not sure where to begin.

Here is where I am looking for help as well; I am feeling confused about the proposed solution above. I think my background is a little lacking on remembering the relation between the summation symbols and the integral symbol in general. I see clearly that we are increasing, but why choose the interval $[k,k-1]$ for example? is it just an arbitrary interval for which we can use the fact that we have an increasing so we can write $x^{q} \le k^{q}$

Then I am confused overall, about the geometric meaning and such. for example how is it true and what is the meaning of writing $$\int_0^n x^q dx = \sum_{k=1}^{n} \int_{k-1}^k x^q dx \leq \sum_{k=1}^{n} \int_{k-1}^k k^q dx $$ . How can I know/see/understand what this is saying is true? and vice versa for the other inequality.

So overall I am looking for any help on that, and with the limit. All is very greatly appreciated and I am trying to understand! Thanks a lot in advance to anyone who is willing to help.

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  • $\begingroup$ The interval $[k-1,k]$ is taken because it has length $1$, hence $$\int_{k-1}^{k}k^qdx=k^q(k-(k-1))=k^q$. For a geometric interpretation, split $[0,n]$ into intervals $[k-1,k]$ and draw bars of height $k^q$ and $(k+1)^q$ and length $1$ for each $k$ and plot $f(x)=x^q$ on the same graph, you should see how these bars are below and above the graph of $f$. For your question regarding $a_n$, note that for large $n$, $a_n$ is close to the integral of $x^q$ from 0 to n considering a partition with intervals of length $1/n^q$. $\endgroup$ – Reveillark Mar 7 '15 at 22:33
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What you have in the first part of your problem is actually two inequalities. The first one looks like a case of the more general

$$\int_0^n f(x)\, dx \leq f(1) + f(2) + \cdots f(n)$$

for an integer $n$ and a function $f$ that is increasing on the interval $[0,n]$. The right-hand side of this inequality is a right-handed Riemann sum of the function $f$ over that interval. The other inequality is a case of

$$ f(0) + f(1) + \cdots f(n+1) \leq \int_0^{n+1} f(x)\, dx$$

for an integer $n+1$ and a function $f$ that is increasing on the interval $[0,n+1]$.The left -hand side of this inequality is a left-handed Riemann sum of the function $f$ over that interval.

As a reminder, the left-handed and right-handed Riemann sums of an increasing function look something like this:

I did not draw $f(x) = x^q$ here because I wanted the first rectangle in the sum to be visible in both cases. But if you make that substitution, then $f(0) = 0$, $f(1) = 1$, and in general $f(k) = k^q$.

Now consider this fact:

$$\int_0^n x^q\, dx = \sum_{k=1}^{n} \int_{k-1}^k x^q\, dx.$$

This is simply a consequence of the fact that you can compute an integral in two pieces: $$\int_a^b f(x)\, dx = \int_a^c f(x)\, dx + \int_c^b f(x)\, dx.$$ Repeat this step $n-1$ times, using a different integer as the new bound of integration each time, and you have a sum of $n$ integrals equal to the original.

The fact $$\sum_{k=1}^{n} \int_{k-1}^k x^q\, dx \leq \sum_{k=1}^{n} \int_{k-1}^k k^q\, dx$$ for an increasing function $f$ is a consequence of the more general fact that if $f(x) \leq g(x)$ when $a \leq x \leq b$, then $$\int_a^b f(x)\, dx \leq \int_a^b g(x)\, dx.$$ In this case you take $f(x) = x^q$ and $g(x) = k^q$. Visually, you can look at the rectangles in the right-hand Riemann sum to see how this works in this particular case: the area of the rectangle between $x = k-1$ and $x=k$ is $f(k)$, and it is greater than the area under the curve $y=f(x)$ in that same interval.

For the second part of the question, notice that

$$a_n= \frac{1 + 2^q + \cdots + n^q}{n^{q+1}}.$$

The numerator indeed looks a lot like the middle part of the inequalities from the first part of the problem. In fact you can use those inequalities to write

$$\frac{1}{n^{q+1}} \int_0^n x^q\, dx \leq \frac{1 + 2^q + \cdots + n^q}{n^{q+1}} \leq \frac{1}{n^{q+1}} \int_0^{n+1}x^{q}dx.$$

You can use the usual formula for $\int x^q\, dx$ to evaluate the definite integrals on both sides. Does it look like the squeeze theorem applies to this problem then?

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  • $\begingroup$ Thank you! So is the way that I can know if it is right hand or left hand as follows; you knew f(1)+f(2)+.. was the right hand limit because you are evaluating from 1 above the bottom limit of integral and the left hand is because you are starting from 0 itself. Really all that changes is f(0) or f(1) with f(n) and f(n+1)? $\endgroup$ – Quality Mar 7 '15 at 23:14
  • $\begingroup$ but only in the case that it is increasing , if not the other way around? $\endgroup$ – Quality Mar 7 '15 at 23:16
  • $\begingroup$ Yes, it's a left-hand sum if you take the value at the left side of each interval. And yes it was important here that the function was increasing, because that meant the left side of each interval gave the smallest value of $f$ on that interval. If the function were decreasing then the right-hand sum would have been on the left of $\leq$ instead of the left-hand sum. $\endgroup$ – David K Mar 8 '15 at 3:14
  • $\begingroup$ That makes sense ! By the way , does 1/p+1 make sense for the limit using squeeze thereom? $\endgroup$ – Quality Mar 9 '15 at 14:53
  • $\begingroup$ You mean $1/(q+1)$? It sounds like you have the right idea. $\endgroup$ – David K Mar 9 '15 at 15:00

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