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.