Answered: Can you further share how to find a…

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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Question

Can you further share how to find a convergent sequence

Expert Solution

Step 1: check whether sequence convergent

Given sequence $open curly brackets X subscript n close curly brackets space$ by $X subscript n equals sum from i equals 1 to n of open parentheses negative 1 close parentheses to the power of i over 2 to the power of i comma n greater or equal than 1$

Here we consider two subsequences $open curly brackets X subscript 2 n end subscript close curly brackets space a n d open curly brackets space X subscript 2 n minus 1 end subscript close curly brackets$ .Noted that A sequence $open curly brackets u subscript n close curly brackets space i n space straight real numbers$ is convergent $left right double arrow$ $open curly brackets u subscript 2 n end subscript close curly brackets space a n d space open curly brackets u subscript 2 n minus 1 end subscript close curly brackets$ are both converges to same limit.

Now observed that $negative X subscript 2 n end subscript equals sum from i equals 1 to 2 n of 1 over 2 to the power of i minus 2 open parentheses 1 over 2 squared plus 1 over 2 to the power of 4 plus... plus 1 over 2 to the power of 2 n end exponent close parentheses equals 1 minus 1 over 2 to the power of 2 n end exponent minus open parentheses 1 half plus 1 over 2 cubed plus... plus 1 over 2 to the power of 2 n minus 1 end exponent close parentheses$

$equals 1 minus 1 over 2 to the power of 2 n end exponent minus 1 half open parentheses 1 plus 1 fourth plus open parentheses 1 fourth close parentheses squared plus.... plus open parentheses 1 fourth close parentheses to the power of n minus 1 end exponent close parentheses equals 1 minus 1 over 2 to the power of 2 n end exponent minus 2 over 3 open parentheses 1 minus 1 over 4 to the power of n close parentheses$

so $negative limit as n rightwards arrow infinity of X subscript 2 n end subscript equals 1 minus 2 over 3 equals 1 third rightwards double arrow limit as n rightwards arrow infinity of X subscript 2 n end subscript equals negative 1 third$ and