<p>Can someone explain to me how to do the following?</p>
<p>What is the value of the summation (from 1 to infinity) of 2^(n+1)/3^n? (and I'm having trouble understanding the tests of convergence...)</p>
<p>and just taylor/Maclaurin series in general... I don't understand it.</p>
<p>VERY MUCH APPRECIATED!!!! THX A MILLION!</p>
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What is the value of the summation (from 1 to infinity) of 2^(n+1)/3^n? (and I'm having trouble understanding the tests of convergence...)
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<p>the summation of 2^(n+1)/3^n is = 2<em>the summation of 2^n/3^n. Notice I pulled a power of two out (since it was 2^n+1), and I put it outside the sigma because it is a constant multiple. Now you are left with a simple geometric series inside the sigma: 2^n/3^n, which = (2/3)^n (law of exponents). The sum of an infinite series a</em>r^n (here a is 1 and r = 2/3) is (1st term)/(1-r). Here the first term (if you list it) is 2/3, and the r is 2/3, so the sum is 2/3/(1-2/3) or 2/3/1/3 = 2. Don't forget about the 2 you pulled out of the sigma, which you now multiply back to get the sum of 4. </p>
<p>Tests of convergence are basically using a limit and an equality to tell you what range x must be in to give you a convergent series as per some rule of convergence (ratio test, etc..) It's really the kind of thing you have to do actually do as opposed to just read about.</p>
<p>Taylor/Maclaurin series is both something you have to understand be able to do. They are series used to calculate the values of complex functions to a desired degree of accuracy. A lot of people have trouble seeing the use for Taylor/Maclaurin series in the age of TI-89s and supercomputers. Please remember that calculus was invented in the 17th century by guys in big white wigs working by light of candle with slide rules and quills. Taylor/Maclaurin series were a lot more useful back then. They are a holdover from antiquity and essentially aren't "needed" any more in order to calculate functions, but are taught because of the high degree of mathematical thinking required to use them. Memorize the series for the elementary functions (cosx, sinx, e^x, lnx, etc), as painful as it may seem, and memorize the general form f(n)(x-c)^n/n! and you will always be right.</p>
<p>I hope i read it right... so you can pull out a two and see it's a geometric series... the summation from 1 to infinity of 2 times 2^n / 3^n.. pull the 2 in front of the sigma and you have 2 sigma 2^n/3^n, or (2/3)^n, a geometric series.. use the formula to find the sum. s = a/1-r, where a is the first term and r is the ratio between the terms. s = 2/3 / 1-2/3 = 2/3 / 1/3 = 2. Multiply that by the two in front of the sigma and you get 4.</p>
<p>Summation (1->Infinity) 2^(n+1)/3^n
Rewrite into: Summation (1->Infinity) 2*(2/3)^n</p>
<p>By Gemo. Series, S=a/(1-r)</p>
<p>In this case, a=2(2/3), r=2/3
S=(4/3) /(1-2/3)=4</p>
<p>Yaylor Series: f(a)+f'(a) (x-a)+f"(a)(x-a)^2 / 2!+f3(a)(x-a)^3 / 3!+f4(a)(x-a)^4 / 4!+........+fn(a)(x-a)^n / n!+...</p>
<p>Thanks so much, I would not have thought of to pull out the 2.</p>
<p>hmm, do you know of any websites that will explain how to compute Taylor series?(or with the solutions for the "elementary ones" that ObsessedAndre referred to?)</p>
<p>bump......</p>
<p>Most calculus textbooks have a page in them somewhere that lists the power series for the elementary functions. </p>
<p>For example, e^x = the summation (from 0 to infinity) of x^n/n! </p>
<p>sinx = the summation (from 0 to infinity) of [(-1)n<em>x^(2n+1)]/(2n+1)!
cosx = the summation (from 0 to infinity) of [(-1)^n</em>x^(2n)]/(2n)!</p>
<p>In my calc class the teacher gave us a sheet that had like a dozen elementary functions on them for us to memorize. I think you can find similar stuff online.</p>
<p>cos(x) about x = 0
cos(0) + -sin(0)x/1! - cos (0)x/2! + sin(0)x/3! + cos(0)x/4!
1 - x/2! + x/4! - x/6! and so on</p>
<p>similiar e^x is its own derivative so its
1 + x/1! + x/2! + x/3!</p>
<p>if you forget these, then just think whats e^0 or cos(0) or sin(0) and that will help to get you started. And in the first one it could be not about 0, but about like pi/4 so it would be
cos(pi/4) + -sin(pi/4)(x-pi/4)/1! - cos (pi/4)(x-pi/4)/2! + sin(pi/4)(x-pi/4)/3! + cos(pi/4)(x-pi/4)/4! I believe.</p>
<p>On google I found some site and it basically shows a graph of a complex function and then shows how if you use say the first 3 terms, what that function looks like compared to the real one, and then as you increase the # of terms it traces more of the graph. Once you get to the end points and beyond of the radius of convergence, it starts deviating a lot, and they can go in totally different directions.</p>