I am sort of confused.

Suppose we are given the series,

$\displaystyle \lim_{n\to\infty}\sum_{k=1}^{n} \frac{k^{99}}{n^{100}}$

How can this be written as an integral, and what would the variable be?

In this series given, which terms are the constants? Is it $n^{100}$??

Wouldn't the above be written as,

$\displaystyle \lim_{n\to\infty} \frac{1}{n^{99}} \cdot \frac{1}{n}\sum_{k=1}^{n}\frac{k^{99}}{1}$

So in the integral, what will be the "respect-to-variable?" Would it be:

$\displaystyle \lim_{n\to\infty} \frac{1}{n^{99}} \int_{0}^{1} k^{99} \text{dk}$

$= \displaystyle \lim_{n\to\infty} \frac{1}{n^{99}} \frac{1}{100}$

But that is wrong as shown here: Limit of a summation, using integrals method

Bottonlinequestion: I am confused about how you write an integral from a SUM. Like what is variable the integral is made with respect to?


$$ \sum_{k=1}^{n} \frac{k^{99}}{n^{100}}=\frac1n\sum_{k=1}^{n}\Bigl(\frac{k}{n}\Bigr)^{99}. $$ This is a Riemann sum for the integral $$ \int_0^1 x^{99}\,dx. $$

  • $\begingroup$ How did the $(k/n)$ turn into an $x$ like in your integral? $\endgroup$ – Amad27 Oct 30 '14 at 12:32
  • $\begingroup$ Do you know what a Riemann sum is? $\endgroup$ – Julián Aguirre Oct 30 '14 at 12:58
  • $\begingroup$ Hi! I do know what a Riemann sum, but the only issue I have is converting the sum into an integral. So how did you do it? $\endgroup$ – Amad27 Oct 30 '14 at 13:36
  • $\begingroup$ Go the other way around. Given the integral, write a Riemann sum on the appropriate partition. $\endgroup$ – Julián Aguirre Oct 30 '14 at 13:43
  • $\begingroup$ Thanks! I'll try here: Using a right hand sum, with $\Delta(x) = \frac{1}{n}$, we must "approximate" the area under $x^{99}$ using $n$ subintervals from $[0, 1]$. Right, the $f(x_i) = \frac{k}{n}$, but how do you get from the sum to the integral? $\endgroup$ – Amad27 Oct 30 '14 at 13:51

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