7
$\begingroup$

Could someone explain how one can replace infinite sum with integral?

Examples (or you may use your own, it doesn't matter as I want to understand principles):

$$\frac{1}{n} \sum_{i=1}^{n} \sin \frac{i-1}{n} \pi$$

$$\frac{1}{n} \sum_{i=1}^{n} \sqrt{1 + \frac{i}{n}}$$

I see that $\frac{1}{n}$ denote partition, so my problem is that I don't know how to figure out the function, hence, the integrand.

$\endgroup$
  • $\begingroup$ Are you looking for something like $\dfrac{\displaystyle \int_{x=0}^{\pi} \sin(x)\, dx }{ \displaystyle \int_{x=0}^{\pi}\, dx}$ and $\dfrac{\displaystyle \int_{x=0}^{1} \sqrt{1+x} \, dx }{ \displaystyle \int_{x=0}^{1} \, dx}$ so in effect finding the mean of the summands or integrand? Or do you want bounds? $\endgroup$ – Henry Jul 10 '15 at 15:02
  • 1
    $\begingroup$ Related : math.stackexchange.com/questions/469885/… $\endgroup$ – lab bhattacharjee Jul 10 '15 at 15:02
  • 2
    $\begingroup$ These are not infinite sums. If one finds $\displaystyle\lim_{n\to\infty} \sum_{i=1}^n a_n$, that can be considered an infinite sum. But $\displaystyle \lim_{n\to\infty} \frac 1 n \sum_{i=1}^n a_n$ is somewhat different from that. Notice that if you push the $1/n$ inside the sum, getting $\displaystyle \lim_{n\to\infty} \sum_{i=1}^n \frac{a_n} n$, then as $n$ increases, one isn't just adding more terms, but actually changing the terms that were already there by dividing them by a larger value of $n$. In $\displaystyle\lim_{n\to\infty} \sum_{i=1}^n a_n$, increasing $n$ merely adds more terms. $\endgroup$ – Michael Hardy Jul 10 '15 at 17:45
  • 1
    $\begingroup$ Bravo, @MichaelHardy, you explained well what I tried to say much more briefly in my answer. $\endgroup$ – Rory Daulton Jul 10 '15 at 17:51
  • $\begingroup$ @RoryDaulton : Thank you. $\endgroup$ – Michael Hardy Jul 10 '15 at 17:57
7
$\begingroup$

The Left-hand Rectangular Approximation Method (LRAM) says that, if $f(x)$ is continuous on $[a,b]$,

$$\int_a^b f(x)\,dx=\lim_{n\to\infty}\frac {b-a}{n}\sum_{i=1}^n f\left(a+\frac{b-a}n (i-1) \right)$$

Comparing that with your first sum, we see that $\ a=0,\ b=\pi, f(x)=\sin x/\pi$. So, the limit of your sum as $n\to\infty$ is

$$\frac{1}{\pi}\int_0^{\pi} \sin x\,dx$$

Note that this is not your sum, as you asked, not even an infinite sum, but a limit of sums.

The Right-hand Rectangular Approximation Method (RRAM) says that, if $f(x)$ is continuous on $[a,b]$,

$$\int_a^b f(x)\,dx=\lim_{n\to\infty}\frac {b-a}{n}\sum_{i=1}^n f\left(a+\frac{b-a}n i \right)$$

Comparing that with your second sum, we see that $f(x)=\sqrt x,\ a=1,\ b=2$. So the limit of your sum as $n\to\infty$ is

$$\int_1^2 \sqrt x\,dx$$

Note: The main difference in these two methods is that LRAM has the term $f(a)$ but not the term $f(b)$ while RRAM has the term $f(b)$ but not the term $f(a)$ in the sum. This is due to the different indices on the sum.

$\endgroup$
  • $\begingroup$ Thank you! Very clear explanation! $\endgroup$ – push Jul 10 '15 at 18:16
  • $\begingroup$ @push: Thanks, I try, especially in my calculus class. Do take note of what MichaelHardy wrote in his comment. $\endgroup$ – Rory Daulton Jul 10 '15 at 18:17
3
$\begingroup$

The basic idea is this:

If $f$ is a positive increasing function then $f(n) \le \int_n^{n+1} f(x) dx \le f(n+1) $.

Similarly, if $f$ is a positive decreasing function then $f(n) \ge \int_n^{n+1} f(x) dx \ge f(n+1) $.

For increasing $f$, $\sum_{i=0}^{n-1} f(i) \le \int_0^{n} f(x) dx \le \sum_{i=0}^{n-1} f(i+1) $ or $\sum_{i=0}^{n} f(i)-f(n) \le \int_0^{n} f(x) dx \le \sum_{i=0}^{n} f(i)-f(0) $ or $f(0) \le \sum_{i=0}^{n} f(i)-\int_0^{n} f(x) dx \le f(n) $.

For a decreasing function, the inequalities are reversed.

From this you can show that for a bounded piecewise monotonic function, the sum converges to the integral.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.