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Real Analysis

Convergent Series Sum

Show that if the series sum(sum sign) to infinity(top) of k=1(bottom) of a_k converges then a_k--->0

Real Analysis Summation

Show that if sum(sum sign) to infinity(top) of k=1(bottom) of a_k=A and sum(sum sign) to infinity(top) of k=1(bottom) of b_k=B, then 1-sum(sum sign) to infinity(top) of k=1(bottom) of ca_k=cA for all c belong to R 2-sum(sum sign) to infinity(top) of k=1(bottom) of (a_k+b_k)=A+B

Real Analysis

Assume a_n and b_n are Cauchy sequences.Use a triangle inequality argument to prove c_n=Absolute value of a_n-b_n is Cauchy.

Real Analysis

Let (a_n) be a bounded sequence and define the set S={x belong to R: x< a_n for infinitely many terms a_n}. show that there exists a subsequence(a_nk) converging to s=sup S

Real Analysis: Cauchy Sequences

Give an example of each of the following or argue that such a request is impossible: 1) A Cauchy sequence that is not monotone. 2) A monotone sequence that is not Cauchy. 3) A Cauchy sequence with a divergent subsequence. 4) An unbounded sequence containing a subsequence that is Cauchy.

Real Analysis : Converging Sequences

Assume (a_n) is a bounded sequence with the property that every convergent subsequence of (a_n) converges to the same limit a belong to R.show that (a_n) must converges to a.

Real Analysis: Sequences

Give an example of each of the following, or argue that such a request is impossible: 1) A sequence that does not contain 0,1 as a term but contains subsequences converging to each of these values. 2) A monotone sequence that diverges but has a convergent subsequence. 3) A sequence that contains subsequences converging to

Real Analysis: Countable Sets and Antichains

Answer the following by establishing 1-1(one to one) correspondence with a set of known cardinality: 1 - Is the set of all functions from{0,1} to N countable or noncountable? 2 - Is the set of all functions from N to {0,1} countable or noncountable? 3 - Given a set B ,a subset A of P(B) is called an antichain if no element of

Real Analysis : Empty Set and Nested Interval Property

Prove that: Intersection to infinity for n=1 (sign of intersection with infinity on top and n=1 in the bottom) of (0,1/n)=empty. (Notice that this demonstrates that the interval in the Nested Interval Property must be closed for the conclusion of the theorm to hold.)

Infinite Series, Convergence and Divergence Test

PLEASE SHOW ALL WORK, STEP-BY-STEP, WITH ALL CORRECT NOTATION. My notation is lousy for some reason the attachment won't come through, so I cut and pasted it below. Determine whether the following Diverge (D), Converge Conditionally (CC), or Converge Absolutely (AC). Give the rationale for each response. Must show all work

Real analysis

Prove: subsequences of a convergent sequences converge to the same limit as the original sequence

Real Analysis

Show that lim inf a_n = lim sup a_n if and only if lim a_n exists

Real Analysis

Show that the sequence defined by y1=1 and y_n+1=4-1/y_n converges and find the limit

Real Analysis

Show that if x_n <= y_n <= z_n for all n belong to N and if lim x_n=limz_n=L then lim y_n=L as well

Real Analysis

A) Show that if (b_n)-->b,then the sequence of absolute values Absolute value of b_n converges to absolute value of b

Real Analysis

Show that limits, if exist, must be unique. In other words, assume lim an=L1 and lim an=L2 and prove that L1=L2

Real Analysis

Let xn(smaller n)>=0 for all n belong to N a) if (xn)-->0, show that(sqrt[xn])-->0 b)if (xn)-->x,show that(sqrt[xn])-->x

Real Analysis

Using the definition of convergence of a sequence show that the following sequences converge to the proposed limit: 1-lim 1/(6n^2+1)=0 2-lim 2/sqrt[n+3]=0 3-lim (3n+1)/(2n+5)=3/2

Real Analysis

1- if A1,A2,A3,...,Am are each countable sets, then the union A1 U A2 U A3...U Am is countable 2- if An is a countable set for each n belong to N,then Un=1(to infinity) An is countable

Real Analysis

Proof: if A subset or equal of B and B is countable, then A is either countable, finite or empty.

Real analysis

Prove that if a is an upper bound for A and if a is also an element of A, then it must be that a=sup A

Real Analysis: Proof Regarding Bounded Subsets

Let A subset or equal of R be bounded above and let c belong to R.Define the sets c+A and cA by c+A={c+a : a belong to A} and cA={ca : a belong to A}. 1-show that sup(c+A)=c+ Sup A 2-if c>=0,show that sup(cA)=cSupA 3-postulate a similar type of statement for sup(cA)for the case c<0.

Real Analysis

Assume that A And B are nonempty, bounded above and satisfy B subset or equal of A. Show that sup B<= sup A