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Convergence of Sequences in Real Analysis
Sajad Ahmad Sheikh
1 Introduction
In real analysis, the concept of convergence plays a fundamental role in under- standing the behavior of sequences. Convergence describes the tendency of a sequence to approach a certain limit as the sequence progresses. It provides a powerful tool for studying the properties and behavior of real numbers.
2 Definition of Convergence
Let (an)∞ n=1 be a sequence of real numbers. We say that the sequence converges to a real number L, denoted by limn→∞ an = L, if for every positive real number ϵ, there exists a positive integer N such that for all n ≥ N , we have |an − L| < ϵ.
2.1 Intuition
The definition of convergence can be understood intuitively as follows: A se- quence (an) converges to a limit L if, as we progress further along the sequence, the terms an get arbitrarily close to L. In other words, for any desired level of closeness (specified by ϵ), we can find a point in the sequence beyond which all terms are within ϵ of L.
3 Importance of Convergence
Convergence is a crucial concept in real analysis due to several reasons.
3.1 Uniqueness of Limits
If a sequence converges, the limit is unique. In other words, if limn→∞ an = L 1 and limn→∞ an = L 2 , then L 1 = L 2. This property allows us to talk about ”the” limit of a sequence rather than multiple possible limits.
3.2 Characterizing Behavior
Convergence provides a way to characterize the behavior of sequences. A se- quence may converge to a fixed limit, diverge to ±∞, or exhibit oscillatory
behavior. By studying the convergence of a sequence, we gain insights into its long-term behavior.
3.3 Computing Limits
Convergence enables us to compute limits of more complicated sequences by re- lating them to simpler sequences whose limits are known. This technique, known as the squeeze theorem, is particularly useful when dealing with sequences that are difficult to evaluate directly.
4 Examples
Let’s consider a few examples to illustrate the concept of convergence.
4.1 Example 1: Geometric Sequence
Consider the sequence (an) =
2 n
n=1.^ We claim that limn→∞^ an^ = 0.^ To prove this, let ϵ > 0 be given. We need to find a positive integer N such that |an − 0 | = |an| < ϵ for all n ≥ N. Note that an = (^21) n. By choosing N = ⌈log 2 (1/ϵ)⌉, we have |an| = (^21) n ≤ (^21) N < ϵ for all n ≥ N. Therefore, limn→∞ 21 n = 0.
4.2 Example 2: Divergent Sequence
Consider the sequence (bn) = (n^2 )∞ n=1. We claim that this sequence diverges to ∞, i.e., limn→∞ bn = ∞. To prove this, let M > 0 be given. We need to find a positive integer N such that bn > M for all n ≥ N. Since bn = n^2 , we can choose N = ⌈
M ⌉. For all n ≥ N , we have bn = n^2 ≥ N 2 ≥ ⌈
M ⌉^2 > M.
Therefore, limn→∞ n^2 = ∞. Consider the sequence (an) =
n
n=1. We claim that limn→∞^ an^ = 0. Solution: To prove that limn→∞ an = 0, we need to show that for any given positive number ϵ, there exists a positive integer N such that |an − 0 | < ϵ whenever n > N. Let’s choose ϵ > 0. We want to find N such that (^1) n − 0 < ϵ for all n > N. Simplifying the inequality, we get (^) n^1 < ϵ, which is equivalent to n > (^1) ϵ. Let N =
ϵ
, where ⌈x⌉ denotes the smallest integer greater than or equal to x. Then, for all n > N , we have n > (^1) ϵ , which implies (^1) n < ϵ. Hence, we have shown that for any ϵ > 0, we can find N such that |an − 0 | < ϵ whenever n > N. Therefore, limn→∞ an = 0.
4.3 Example 3
Consider the sequence (bn) =
(^) n+ n
n=1. We claim that limn→∞^ bn^ = 1. Solution: To prove that limn→∞ bn = 1, we need to show that for any given positive number ϵ, there exists a positive integer N such that |bn − 1 | < ϵ whenever n > N.
To prove this formally, we need to show that for any given positive number ϵ, there exists a positive integer N such that |dn − 0 | < ϵ whenever n > N.
Let’s choose ϵ > 0. We want to find N such that (−1)
n n −^0 < ϵ^ for all n > N. Simplifying the inequality, we get (^1) n < ϵ, which is equivalent to n > (^1) ϵ. Let N =
ϵ
, where ⌈x⌉ denotes the smallest integer greater than or equal to x. Then, for all n > N , we have n > (^1) ϵ , which implies (^1) n < ϵ. Hence, we have shown that for any ϵ > 0, we can find N such that |dn − 0 | < ϵ whenever n > N. Therefore, limn→∞ dn = 0.
4.6 Example 5
Consider the sequence (en) =
n^2 n+
n=
. We want to determine if the sequence
converges or diverges. Solution: To analyze the convergence of the sequence, we will investigate the limit of the sequence as n approaches infinity. Taking the limit, we have:
lim n→∞ en = lim n→∞
n^2 n + 1
To simplify the expression, we can divide both the numerator and the de- nominator by n:
lim n→∞
n^2 n + 1
= lim n→∞
n 1 + (^1) n
As n approaches infinity, the term (^) n^1 approaches 0. Therefore, we have:
lim n→∞
n 1 + (^) n^1
Since the limit of the sequence is ∞, we can conclude that the sequence (en) diverges.
5 Exercises
5.1 Exercise 1
Consider the sequence (an) =
n^2 2 n
n=
. Determine if the sequence converges or
diverges. If it converges, find its limit.
5.2 Exercise 2
Consider the sequence (bn) =
(^) n! nn
n=1. Determine if the sequence converges or diverges. If it converges, find its limit.
5.3 Exercise 3
Let (cn) be a sequence defined by c 1 = 1 and cn+1 =
2 + cn for n ≥ 1. Determine if the sequence converges or diverges. If it converges, find its limit.
5.4 Exercise 4
Consider the sequence (dn) =
(^2) n n!
n=1. Determine if the sequence converges or diverges. If it converges, find its limit.
5.5 Exercise 5
The Fibonacci sequence is defined recursively as F 1 = 1, F 2 = 1, and Fn+2 =
Fn+1 + Fn for n ≥ 1. Prove that the sequence
Fn+ Fn
n=
converges. If it
converges, find its limit.
5.6 Exercise 6
Consider the sequence (en) =
sin
(^) nπ 2
n=1.^ Determine if the sequence con- verges or diverges. If it converges, find its limit.
5.7 Exercise 7
Let (fn) be a sequence defined recursively by f 1 = 1, f 2 = 2, and fn+2 = fn+1 +
fn for n ≥ 1. Determine if the sequence converges or diverges. If it converges, find its limit.
5.8 Exercise 8
Consider the sequence (gn) =
(^2) n+3n 4 n
n=1. Determine if the sequence converges or diverges. If it converges, find its limit.
5.9 Exercise 9
Let (hn) be a sequence defined by h 1 = 0 and hn+1 =
hn + 2 for n ≥ 1. Determine if the sequence converges or diverges. If it converges, find its limit.
5.10 Exercise 10
Consider the sequence (in) =
(^) n! en
n=1. Determine if the sequence converges or diverges. If it converges, find its limit.
