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So the example is in the page 139 of Chapter V, Definite Integrals

Form the integral sum $S_n$ for the function

$$ f(x) = 1 + x $$

on the interval $[1,10]$ by dividing the interval into n equal parts and choosing points $\xi_i $ that coincide with the left end-points of the subintervals $[x_i, x_{i+1}]$. What is the $$ \lim_{n\to \infty} S_n $$ equal to?

Solution:

Here, $$ \Delta x_i = \frac{10-1}{n} = \frac{9}{n}$$

and

$$ \xi_i = x_i = x_0 + i\Delta x_i$$

Whence, $$ \xi_i = 1 + \frac{9i}{n} $$

So, $$S_n = \sum_{i=0}^{n-1} f(\xi_i)\Delta x_i = \sum_{i=0}^{n-1}\Bigl(2 + \frac{9i}{n}\Bigr)\frac{9}{n} $$

And then, the next passage is

$$ =\frac {18}{n} n + \frac {81}{n^2} (0+1+...+n-1)= $$

First, I did not understand why the $n$ appeared in the first term of the sum.

$$18+\frac{81}{n^2}\frac{n(n-1)}{2}$$

Next, I did not understand why

$$\frac{n(n-1)}{2}=(0+1+...+n-1)$$

Could anyone help me understood what happened in these 2 passages?

highentropy1915
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3 Answers3

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The second one is the sum of the first n terms of the arithmetic progression $\{ a_0,a_1,\cdots,a_{n-1} \}$ of ratio 1 and $a_0=0$. So we have the formula that is proved by induction: $$ S_n=\dfrac{n(a_{n-1}+a_{0})}{2}. $$ The first one you can see more easily if you develop the sum.

Math_Hater
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  • I don't get why $$\sum_{i=0}^{n-1}1 = n$$

    And yes, my math basis are very very poor

     – highentropy1915 Sep 07 '19 at 00:56
  • The letter below sigma ($i$) indicates the variable that will go from $0$ to $n-1$. In the sum you wrote that letter doesn't appear. So this can't be equal to $n$. Instead, You could have $$\sum_{i=0}^{n-1}i=0+1+2+3+\cdots+(n-2)+(n-1)=\dfrac{n((n-1)+0)}{2}=\dfrac{n(n-1)}{2}.$$ In the formula we have the product of the number of terms in the sum ($n$) with the sum of the last term and the first. All that over $2$. – Math_Hater Sep 07 '19 at 01:29
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The summation $$1+2+\cdots +n-1={n(n-1)\over 2}$$can be proved by induction very easily. For the other one we have $$S_n{=\sum_{i=0}^{n-1}(2+{9i\over n}){9\over n}\\=\sum_{i=0}^{n-1}{18\over n}+{18i\over n^2}\\={18\over n}\sum_{i=0}^{n-1}1+\sum_{i=0}^{n-1}{18i\over n^2}\\={18\over n}\cdot n+{18\over n^2}\sum_{i=0}^{n-1}i\\=18+{18\over n^2}\cdot{n(n-1)\over 2}}$$

Bernard
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Mostafa Ayaz
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  • I think I'm asking stupid questions but it seems my mathematical basis are just like crap. I don't know how to prove things by induction or any other techniques. I'm going to see how can I do it.

    Another thing I don't get is why $$\sum_{i=0}^{n-1}1 = n$$

     – highentropy1915 Sep 07 '19 at 00:47
  • For example in this case you only need to prove that $${n(n-1)\over 2}+n={n(n+1)\over 2}$$(simple, huh?!) Also with the $\sum_{i=0}^{n-1}1$, you are summing up a same number $n$ times, so it will be $n\times 1=n$ – Mostafa Ayaz Sep 07 '19 at 07:56
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$$\displaystyle\sum_{i=0}^{n-1}\frac{18}{n}=\frac{18}{n}\displaystyle\sum_{i=0}^{n-1}1$$ $$=\frac{18}{n}\underbrace{(1+1+\ldots+1)}_\text{n times}$$ $$=\frac{18}{n}n$$

Joel Pereira
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