V8(4n)!(1103 + 26390n)9801 · 3964m (n!)42. Now consider the series >This series was discovered by the extraordinary Indiann=0mathematician Srinivasa Ramanujan (1887-1920). This series converges to1(You do NOT need to prove that,and it is much more difficult than finding the sum of the series in problem 1.) This series has been used to computeT to over 17 million digits (which was a world record at the time).V8(4n)!(1103 + 26390n)9801 · 3964n (n!)400(a)Use any test for convergence/divergence to show that the seriesn=0converges.kV8(4n)!(1103 + 26390n)9801 · 3964n (n!)4(b)The partial sums for this series are Sk = >n=0Use a calculator to evaluate and , and write down as many digits as your calculator can display. Howmany digits are the same as the digits of T?Note: T 2 3.1415926535 8979323846 2643383279...

(a) Given series, ∑n=0∞an = ∑n=0∞8(4n)!(1103+26390n)9801·3964n(n!)4 Here, an =…

Answered: 2. Now consider the series v8(4n)!(1103…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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1 (4) Find the sum of the telescoping series 2n=1 244n+3
(Babylonian Method). All numerical answers should be rounded to 6-digit floating-point numbers. (0) If A > 0 is a positive number, then given a real number xo, the sequence (xn) defined recursively by where n > 0 is a natural number, converges to √A, that is, provided that the initial term xo is chosen 'not too badly'. In practice, calculation of terms xn is carried out till this or that stopping criterion is triggered. (i) Let (xn) be the sequence defined recursively by (ii) Generate the terms and x0 = 94 (as it could be guessed from part (0), the sequence (xn) converges to √946). x() = Xn+1= = = (x++) ₁ (n ≥ 0), x1 = x2 = of the sequence (xn) till you find the term x N satisfying the condition x3 = fl(xN) = fl(xN-I (our stopping criterion), where fl(z) denotes the result of rounding of a real number z to a 6-digit floating-point number. Along the way, once the k-th term xk is generated, enter in the corresponding input field the result fl(xk) of rounding of xk to a 6-digit…
3. The series 1 (4n) !(1103 + 26390n) 9801 (n!)*3964n was discovered by Ramanujan in the early 1900s. William Gosper, Jr., used this series to calculate a to an accuracy of more than 17 million digits in the mid-1980s. At the time, that was a world record. Since that time, this series and others by Ramanujan have led mathematicians to find many other series representations for T and 1/т. a. Prove that this series converges.
12) There is one special “family” of series for which, if the series converges, we can determine the exact value of the series. This is the family where it really matters that we know the bounds of the summation. What family is this?
compare the series (1+x)-3=1-3x+6x2-10x3........ with ex= 1+x +x2/2 +x3/6............ Are they both convergent? How can you prove this?
For problem 2 part 2: To get your comparison series for LCT, choose the series with leading term over leading term. So you want the series of sqrt{n} over n "leading term over leading term" then go from there
I need help figuring out what process to use to complete this problem. If p, q, and r are all positive then what is the collection of all positive values p, q, and r that cause all three series in the picture to converge? p>0, q>0, r>1 p>1, q>1, r>0 p>1, q>1, r>1 p>1, q>0, r>0
1. In this problem, you'll look at the series > 1 )J: One can visualize this series as area, by drawing a k5/4 * k=10 box of area 103/4, a box of area 134 and so on. 115/4 (a) There are two ways to draw the series (as boxes) in question. Draw each of them together with the graph y= 34 on two different graphs. x5/4 (b) Use the picture to write an inequality comparing >) 1 to an improper integral. Check the k5/4 k=10 bounds of your improper integral carefully, and be sure you are comparing on the same domain. (c) Is the improper integral you wrote in (b) convergent or divergent? Explain. (d) Using the inequality you wrote in (b) and your answer to (c), what (if anything) can you conclude 1 :? about the convergence of 2 k5/4 k=10 (e) Repeat (b)-(d) with your second picture.
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