**Question 3**The series $1 - \frac{4}{9} + \frac{16}{81} - \frac{64}{729} + \ldots$ is a [blank] series with [blank]. Therefore, it [blank].**Question 4**The geometric series $\sum_{n=1}^{\infty} \left(\frac{e}{3}\right)^n$ converges and has sum $S =$ [blank].

Part A- Concept Used: (Ratio test)- If limn→∞an+1an=L then If L<1 then series convergent If…

Answered: QUESTION 3 16 The series 1- 9. 64 +- is…

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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Step 1

Part A-

Concept Used: (Ratio test)-

If $\lim_{n \to \infty} |\frac{a_{n + 1}}{a_{n}}| = L t h e n$

If $L < 1$ then series convergent

If $L > 1$ then series divergent

If $L = 1$ then series inconclusive

Given:

Given series is

$1 - \frac{4}{9} + \frac{16}{81} - \frac{64}{729} + . . . . . .$

Calculation:

Given series is

$1 - \frac{4}{9} + \frac{16}{81} - \frac{64}{729} + . . . . . . \Rightarrow \sum_{n = 0}^{\infty} {(- \frac{4}{3})}^{n}$

This is Geometric series with the first term $b = 1$ and common ratio $q = - \frac{4}{3}$

Now $|q| > 1$ By using Ratio test

$\begin{array}{rcl} \lim_{n \to \infty} |\frac{a_{n + 1}}{a_{n}}| & = & \lim_{n \to \infty} |{(- \frac{4}{3})}^{n + 1} \times {(- \frac{4}{3})}^{n}| \\ = & \lim_{n \to \infty} |\frac{- 4 {(- 4)}^{n}}{3 {(3)}^{n}} \times \frac{{(3)}^{n}}{{(- 4)}^{n}}| \\ = & \frac{4}{3} > 1 \end{array}$

Hence Series is divergent.

Answer:

Given series $1 - \frac{4}{9} + \frac{16}{81} - \frac{64}{729} + . . . . . .$ geometric series with the first term $b = 1$ and common ratio $q = - \frac{4}{3}$ , $|q| > 1$

Therefore By ratio test it is divergent.

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