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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?

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Amad27
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1 Answers1

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

Julián Aguirre
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  • How did the $(k/n)$ turn into an $x$ like in your integral? – Amad27 Oct 30 '14 at 12:32
  • Do you know what a Riemann sum is? – Julián Aguirre Oct 30 '14 at 12:58
  • 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? – Amad27 Oct 30 '14 at 13:36
  • Go the other way around. Given the integral, write a Riemann sum on the appropriate partition. – Julián Aguirre Oct 30 '14 at 13:43
  • 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? – Amad27 Oct 30 '14 at 13:51
  • Practice! You have to recognize from the form of the sum the right function to put under the integral sign. – Julián Aguirre Oct 30 '14 at 13:53
  • Thanks! Also, how did you choose the limits of the integral? I know $\Delta(x) = \frac{b-a}{n}$, but $b - a = 1$ could fit infinitely many values, $(b, a) = (1, 0), (2, 1), (3, 2).... (n, n-1)$ ? – Amad27 Oct 30 '14 at 13:54
  • $k/n$ goes from $1/n\approx0$ to $n/n=1$. – Julián Aguirre Oct 30 '14 at 14:33
  • Hi! The only issue is the approximately $0$ and not exactly $=0$ and because $n/n = 1$, that shows the height of the function in the last subinterval $=1$, which is only possible at $x = 1$ right? – Amad27 Oct 30 '14 at 14:36
  • I am sorry, I think that you should do some thinking by yourself. – Julián Aguirre Oct 30 '14 at 14:39