Sequences.A. Is it true that if a sequence (un)nEN in R is convergent if |un+1 − un| → 0. Justify your answerB. Is it true that if a sequence (un)nEN in R is convergent if |un+1 − un|≤1/(3n). Justify your answerC. Show that a sequence (un)nEN in R is convergent if its two subsequences (u2n)nEN and (u2n+1)nEN are convengent to the same limit.

Name: Sequences.A. Is it true that if a sequence (un)nEN in R is convergent
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Description: Sequences.A. Is it true that if a sequence (un)nEN in R is convergent if |un+1 − un| → 0. Justify your answerB. Is it true that if a sequence (un)nEN in R is convergent if |un+1 − un|≤1/(3n). Justify your answerC. Show that a sequence (un)nEN in R is

C. Let, ε>0 Let, limn→∞un=l=limn→∞un+1, then there exists natural numbers k1,k2 such that…

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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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Sequences.

A. Is it true that if a sequence (u_n)_nEN in R is convergent if |u_n+1 − u_n| → 0. Justify your answer

B. Is it true that if a sequence (u_n)_nEN in R is convergent if |u_n+1 − u_n|≤1/(3ⁿ). Justify your answer

C. Show that a sequence (u_n)_nEN in R is convergent if its two subsequences (u_2n)_nEN and (u₂_n+1)_nEN are convengent to the same limit.

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Step 1

C. Let, $ε > 0$

Let, $\lim_{n \to \infty} u_{n} = l = \lim_{n \to \infty} u_{n + 1}$ , then there exists natural numbers $k_{1}, k_{2}$ such that

$|u_{2 n} - l| < ε \forall n \geq k_{1} |u_{2 n + 1} - l| < ε \forall n \geq k_{2}$

Let, $k = m a x \{k_{1}, k_{2}\}$

Then, $k$ is a natural number and for all $n \geq k, l - ε < u_{2 n} < l + ε$ and $l - ε < u_{2 n + 1} < l + ε$

That means , $l - ε < u_{n} < l + ε \forall n \geq 2 k - 1$

As $2 k - 1$ is a natural number, it follows that $\lim_{n \to \infty} u_{n} = l$

Then, it is proved that a sequence ${(u_{n})}_{n \in ℕ}$ in $ℝ$ is convergent if its two sub sequences ${(u_{2 n})}_{n \in ℕ}$ and ${(u_{2 n + 1})}_{n \in ℕ}$ are convergent to the same limit.

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