15 ???. 2013 ?. By comparison test the series ? sn converges
Let (sn) be a sequence that converges Assume all sn = 0 and the limit L = lim ... sn. . . of positive real numbers and suppose that limrn = L < 1.
b) Suppose that (sn) and (tn) are sequences such that
# 5 Suppose {sn} and {tn} are real sequences and limn?? sn = s. Show limksn = ks and lim(k + sn) = k + s for all kR. a)
Any convergent sequence is bounded. Proof: Suppose that sn ? s as n ? ?. Taking ? = 1 in the definition of convergence gives that there exists a number N
We first show that one sequence (sn) can not have two different limits. Suppose sn ? s and sn ? t. Let ? > 0. Then ?. 2. > 0. Since sn ? s by definition
Suppose that (sn + tn) converges to a and (tn) converges to b. Let c) Let (sn) be a convergent sequence and let (snk ) and (smk ) be two subsequences of ...
converges to a sum S ? R if the sequence (Sn) of partial sums. Sn = Suppose that (an) is a sequence of real numbers and let r = lim sup.
Let {an} be a bounded sequence such that every convergent subsequence of {an} This is an example of a telescoping series. Since. ? lim n=1 sn = ? lim.
Any convergent sequence is bounded Proof: Suppose that sn ? s as n ? ? Taking ? = 1 in the definition of convergence gives that there exists a number N
It follows that limn?? sn = 1 Therefore the series converges and its sum is 1 The following results can be easily derived from the above definition
#1 Use the definition of convergence to show limn?? 3n+1 n+2 = 3 Let ? > 0 5 Suppose {sn} and {tn} are real sequences and limn?? sn = s
Every convergent sequence is bounded Proof Let (sn) be a sequence that converges to s ? R Applying the definition to ? = 1 we see
8 5a) Claim: Suppose that (an) (bn) and (sn) are three sequences and that an ? sn ? bn for all n ? N If liman = limbn = s then (sn) converges and limsn =
Math 3150 Fall 2015 HW2 Solutions Problem 1 Let (sn) be a sequence that converges (a) Show that if sn ? a for all but finitely many n then limsn ? a
Prove that convergence of {sn} implies convergence of {sn} Is the con- verse true? (Assume we are working in Rk) Solution Let ab ? Rk Then by the
Let (sn) be a sequence of real numbers and let s ? R We say that (sn) We say that (sn) converges if it has a limit and that it diverges otherwise
Let {an} be a bounded sequence such that every convergent subsequence of {an} has a limit L Prove that limn?? an = L Solution Method 1: Note that La = {L}
Suppose S is closed and suppose (sn) is a conv sep of points in So Also suppose lim sn S ? Sry is a convergent sequence in S\{x}