I need to solve #9 a by using the negation of convergence. 9. Consider the sum S1-141-141-141- On one hand, it seems as though S=(1-1) + (1-1)+(1-1) +0+0+0+0. On the other hand, it seems as thoughS=1+ (-1+1)+(-141)+ 14040401. This seems contradictory as 0 / 1,Work through the following problems to determine what is happening here.a. Consider the sequence of partial sums (2k-1) = (a₁ + a₂ + az + + a)(1,0,1,0,1,0,....), Determine whether this sequence converges or diverges. Then proveyour result,b. What does your answer for part b imply in terms of the infinite sum S=2-19}?10. Prove that any integer that is a perfect cube is no more than one integer away from a multiple of9,

Answered: Consider the sum (1-1) + (1-1) + (1-1)…

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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Part a: Convergence or Divergence of the Partial Sums Sequence

The sequence of partial sums is given by:

${\sum_{k = 1}^{i} a_{i}} = {a_{1} + a_{2} + a_{3} + \dots + a_{i}} = {1, 0, 1, 0, 1, 0, \dots}$

To determine whether this sequence converges or diverges, we need to check if the sequence of partial sums approaches a finite limit as $i$ approaches infinity.

Looking at the sequence, we can see that it does not approach a finite limit; instead, it oscillates between 1 and 0 indefinitely. Therefore, we can say that the sequence diverges.

Proof:

To prove this, we can consider two subsequences: one consisting of the partial sums at even indices and the other consisting of the partial sums at odd indices.

The subsequence at even indices is $left parenthesis 0 comma 0 comma 0 comma horizontal ellipsis right curly bracket$ , which is a constant sequence and therefore converges to 0.
The subsequence at odd indices is $left curly bracket 1 comma 1 comma 1 comma horizontal ellipsis right curly bracket$ , which is also a constant sequence and therefore converges to 1.

Since the two subsequences converge to different limits, we can conclude that the original sequence diverges.