Quod Erat Demonstrandum

2007/09/24

[AL][PM] Sum an Infinite Series by Definite Integrals

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 6:39 下午

Using definite integrals to find the sum of an infinite series may be regarded as an application of definite integral. In the text book, we may find the following relation

\lim_{n\rightarrow \infty}\sum_{k = 1}^{n}f(\frac{k}{n})\frac{1}{n} = \int_0^1f(x)dx —— (*)

Actually, it is NOT the unique way of expressing \int_0^1f(x)dx. A more general way is something like

\int_0^1f(x)dx = \lim_{w \rightarrow 0}\sum_{i=1}^{n}f(w_i)\delta x_i

where {x_0 = 0,x_1,x_2,...,x_n = 1} is a partition of [0,1], \delta x_i = x_i - x_{i-1}, w_i \in [x_{i-1} , x_i] and w = \max_{1 \le i \le n}\delta x_i.

But this rigourous definition of Riemann Sum is not a good way of computation of the sum of infinite series, I need students to memorize (*). And I’d like to present the technique in applying (*).

e.g. 1 Evaluate \lim_{n \rightarrow \infty}(\frac{1}{\sqrt{n^2 + n}} + \frac{1}{\sqrt{n^2 + 2n}} + ... + \frac{1}{\sqrt{n^2 + 2n^2}})

STEP 1 : Taking out the common factor \frac{1}{n}
STEP 2 : Express in summation form (optional), try to write the expression involving the term \frac{k}{n}
STEP 3 : Convert the summation into definite integral by regarding

\sum as \int
\frac{1}{n} as dx
\frac{k}{n} as x
The lower limit of the integral = \lim_{n \rightarrow \infty} \frac{lower\:limit\:of\:k}{n}
The upper limit of the integral = \lim_{n \rightarrow \infty} \frac{upper\:limit\:of\:k}{n}

OK, let’s do e.g. 1

\lim_{n \rightarrow \infty}(\frac{1}{\sqrt{n^2 + n}} + \frac{1}{\sqrt{n^2 + 2n}} + ... + \frac{1}{\sqrt{n^2 + 2n^2}})
= \lim_{n \rightarrow \infty}\frac{1}{n}(\frac{1}{\sqrt{1 + \frac{1}{n}}} + \frac{1}{\sqrt{1 + \frac{2}{n}}} + ... + \frac{1}{\sqrt{1 + \frac{2n}{n}}}) [STEP 1 : taking out \frac{1}{n}]
= \lim_{n \rightarrow \infty}\sum_{k = 1}^{2n}(\frac{1}{\sqrt{1 + \frac{k}{n}}})\frac{1}{n} [STEP 2 : express in summation form, obtaining terms in \frac{k}{n}]
= \int_0^2\frac{1}{\sqrt{1 + x}}dx

This is the STEP 3.

\sum becomes \int
\frac{1}{n} becomes dx
\frac{k}{n} becomes x
The lower limit of k = 1, hence \lim_{n \rightarrow \infty}\frac{k}{n} = \lim_{n \rightarrow \infty}\frac{1}{n} = 0
The upper limit of k = 2n, hence \lim_{n \rightarrow \infty}\frac{k}{n} = \lim_{n \rightarrow \infty}\frac{2n}{n} = 2
Therefore, the lower and the upper limit of x are 0 and 2 respectively.
Thus,
\lim_{n \rightarrow \infty}(\frac{1}{\sqrt{n^2 + n}} + \frac{1}{\sqrt{n^2 + 2n}} + ... + \frac{1}{\sqrt{n^2 + 2n^2}})
= \int_0^2\frac{1}{\sqrt{1 + x}}dx
= 2\sqrt{3} - 2

e.g. 2 Evaluate \lim_{n \rightarrow \infty}\sum_{k = 1}^{3n}\frac{1}{n}\sin(\pi + \frac{k\pi}{2n})

Here, STEP 1 & 2 are done. Go to STEP 3 directly.

\sum becomes \int
\frac{1}{n} becomes dx
\frac{k}{n} becomes x
The lower limit of k = 1, hence \lim_{n \rightarrow \infty}\frac{k}{n} = \lim_{n \rightarrow \infty}\frac{1}{n} = 0
The upper limit of k = 3n, hence \lim_{n \rightarrow \infty}\frac{k}{n} = \lim_{n \rightarrow \infty}\frac{3n}{n} = 3
Therefore, the lower and the upper limit of x are 0 and 3 respectively.

Thus,

\lim_{n \rightarrow \infty}\sum_{k = 1}^{3n}\frac{1}{n}\sin(\pi + \frac{k\pi}{2n})
= \int_0^3\sin(\pi + \frac{\pi x}{2})dx
= -\frac{2}{\pi}

After some practice, you may write down the integrals by using your naked eyes. That is, you should have no problem to obtain the following

\lim_{n \rightarrow \infty}\sum_{k = 1}^{2n}\frac{1}{n}\ln(1 + \frac{k^2}{n^2}) = \int_0^2\ln(1 + x^2)dx

Um…Enough? How to evaluate the following?

\lim_{n \rightarrow \infty}(\frac{1}{n + 1} + \frac{1}{n + 3} + \frac{1}{n + 5} + ... + \frac{1}{3n + 1})

廣告

10 則迴響 »

  1. Erm, the first time to visit your site… what should I say… I just feel that it is where you find the crazy equations, and I dunno a word (or should I say symbo?). After all, you’re a professor. I hope I’ve not disturb(?) in any way, with respect.
    Student,
    2V,
    32.
    P.S. don’t give me a bad pt. knowing I’ve been in this blog. ^^ I know you’re busy, so no need to have a reply, uh.

    迴響 由 Edmund — 2007/09/25 @ 11:18 下午 | 回應

  2. Thank you Edmund, you are the first student leaving message here in my blog. So happy! Just having a glance at your blog, I’m impressed by your writing in English. You’re a F.2 boy ar!

    迴響 由 johnmayhk — 2007/09/26 @ 1:33 上午 | 回應

  3. Hi! I haven’t visited your forum for a long time and have just discovered this blog of you. I guess I will visit here often.

    迴響 由 Soarer — 2007/10/02 @ 1:46 下午 | 回應

  4. Soarer, thank you! I guess, in the near future, I’ll call you mathematician. ^_^

    迴響 由 johnmayhk — 2007/10/02 @ 3:38 下午 | 回應

  5. Erm, do you have a nice Mid-Autumn Festival?
    By the way sorry to start up that mess with ‘you haven’t taught it’…
    This quiz is quite a challenge, for me… so is the regular test easier or harder?
    Good day

    迴響 由 Edmund — 2007/10/04 @ 8:54 下午 | 回應

  6. Edmund, it’s nice to be with my family during the holiday.

    When discussing the Q.7, many students couldn’t obtain full mark because they all mistook that AD//BC at the beginning. Studens should, at least, gave some comments or steps to show that they really knew that AD may not be parallel to BC. But they just shouted out that I had not taught the reasoning “sides opp. equal angles" and turned a blind eye to the most important matter. As I replied in the lesson. Even they don’t know the “reason", they CAN solve the problem in another way, but they really need to show (not use) the fact that angle B equals to angle C. For the regular test, I think the questions are easier, but for few students, it is unlikely to get a pass easily.

    迴響 由 johnmayhk — 2007/10/05 @ 12:44 下午 | 回應

  7. Oops, Edmund, you are the last person to leave a message to my old guestbook! The service will terminate forever! Oh my. And I tell you, Rico, she is a beautiful English teacher, it seems that she is not very good at Mathematics (I guess ^_^).

    迴響 由 johnmayhk — 2007/10/05 @ 12:49 下午 | 回應

  8. Amazing Blog…thanks !!!!

    迴響 由 priyam — 2013/05/05 @ 2:10 上午 | 回應

  9. Hi there,

    How about lim (1/n) sigma (k/n) = Sxdx from 0 to 1 = 1/2 ???

    迴響 由 Passer by — 2013/11/29 @ 8:48 下午 | 回應


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