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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(2) An interesting fact is that the Root Test (stated at the end of the previous problem) is stronger than the Ratio Test. Proving this fact would require showing two things: First, one can show that any time the Ratio Test limit exists and equals some L then the Root Test limit also exists and equals the same L. Second, one can find examples of series where the Root Test works but the Ratio Test does not. Your task in this problem is to do the second thing above. To be more specific, find an example of a series a, such that • lim Vlan[ exists and equals a number less than 1, meaning the series converges absolutely by the Root Test, and • lim does not exist. Give a brief explanation why your example satisfics the above two properties.
1 (4) Find the sum of the telescoping series 2n=1 244n+3
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
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.
Hello, Can someone show to solve problem 4 using the root test? Thank you!
4. Does the series formed by subtracting the series 1 from the series 2n - 1 n=1 n3D1 converge? Give reasons for your answer.
PART V: (a) Show that the following series converges conditionally. E-1)+1_n² n° +2 (b) Find an estimate for the error in approximating the series by the sum of its first five terms, then find an estimate for the sum of the series.

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