Purchase Solution

Proof : Sequences and Supremum

Not what you're looking for?

Ask Custom Question

Suppose that the sequences {a_n}_n is bounded above and lim(b_n) exists.

a) Prove that for all e>0 there is an N st that for all n>=N
sup{a_k:k>=n} + b_n <=sup{a_k + b_k: k>=n} + e.

b) Use this to conclude

limsup(a_n) + lim (b_n) <= limsup (a_n+b_n)

(all limits are n ---> infinity).

Purchase this Solution
Solution Summary

Sequences and supremum are investigated in the solution.

Solution Preview

Define sup(a_n)=sup{a_k:k>=n}

Proof:
(a) Since lim(b_n) exist, we can suppose lim(b_n)=b. Then for all e>0, there is an N such that for ...

Purchase this Solution
Free BrainMass Quizzes
Geometry - Real Life Application Problems

Understanding of how geometry applies to in real-world contexts

Exponential Expressions

In this quiz, you will have a chance to practice basic terminology of exponential expressions and how to evaluate them.

Probability Quiz

Some questions on probability

Multiplying Complex Numbers

This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form.

Solving quadratic inequalities

This quiz test you on how well you are familiar with solving quadratic inequalities.

View More Free Quizzes
Related BrainMass Solutions

  • Sequences and Limit Superior

    113189 Sequences and Limit Superior Suppose that xn x and the sequence (yn) is bounded.

  • Real Analysis of Upper Bounds

    a proof regarding the upper bound and the supremum.

  • Continuity and Sequential Limits

    Hence f has supremum. Similarly we can prove it for infimum m. Continuity and Sequential Limits are investigated.

  • Proof: Upper and lower limits

    Proof. 1) By the definition of infimum and supremum, we know . So we have 2) If the limit of only exists when , say it equals A, ie., . Since , . Obviously is increasing and , so exists and equal to A.

  • Real analysis proofs

    There are two real analysis proofs here, one regarding unions and supremum and one regarding measurable functions.

  • Sequence of Functions and Mean Value Theorem

    Proof: Since Sup|f'_n(x)|<=M, then for any n and any x on [a,b], we have |f'_n(x)|<=M. Since f_n is differentiable on (a,b), according to the Mean Value Theorem, we have f_n(y)-f_n(x)=f'_n(z)(y-x), where z is between x and y.

  • Proof: Uniformly Continuous

    84783 Proof: Uniformly Continuous The following theorem could be used to write the proof.

  • Bounded Real Sequences and Metrics

    Remark: From the above proof, maybe the statement (2) is not so obvious. If (ak) and (bk) are both increasing or decreasing sequences, the inequality (2) becomes equality.

  • Riemand integrable

    if S has or as the only one limit point, then Proof: Let M=supremum of f(x) on [a, b] and m=infimum of f(x) on [a, b]. Since f is discontinuous on [a, b] . Let be given. Choose a point such that .

View More Related Solutions