(b) Show that if L > 1, then limsn = +∞ Solution (a) Define the sequence rn = \ \ \ sn+1
MATH fa hmw solutions
(c) Prove that if limsn and limtn exist, then limsn ≤ limtn Solution (a) Let M > 0 Since sn → +∞ (b) Show that if L > 1, then limsn = +∞ Hint: Apply (a) to the
HW solution
sn ) ≤ lim m→∞ ( inf n>m tn ) = lim inf tn A similar proof works for lim sup Problem 7 (12 2) Prove that lim supsn = 0 if and only if lim sn = 0 Solution If lim
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If (sn) converges to s then we say that s is the limit of (sn) and write s = limn sn, or Proof (a) (⇐) If (sn) is a sequence in S with limit x, and if ϵ > 0 is given, then
t ntsumm
a) a problem from the previous sheet b) the Squeeze Theorem 4) Prove that if sn ≤ tn for all n and lim n→∞ sn = ∞, then lim n→∞ tn = ∞ 5) Use the Squeeze
problemsheet
lim bn = s Prove lim sn = s This is called the “squeeze lemma ” (b) Suppose (sn) and (tn) are (a) Show that if sn ≥ a for all but finitely many n, then lim sn ≥ a
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Section 12 12 2 Prove limsup sn = 0 if and only if lim sn = 0 12 4 Show limsup( sn + tn) ≤ lim sup sn + lim sup tn for bounded sequences (sn) and (tn) Hint
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8 4) Claim: If (tn) is a bounded sequence and (sn) is a sequence such that limsn = 0, then (Alternatively, we could simply prove directly that lim−tn = 0 ) Then
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These examples should suggest a result: Theorem 2 20 Let (sn) be a sequence Then lim inf sn = lim sup sn if and only if lim sn = s for
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(b) Show that if L > 1 then lim
Now since any sequence (sn) with a limit is either bounded above or bounded below we conclude that (an) has no limit if a < ?1. 9.16 (a) Prove lim n4+8n n2+9.
8.4) Claim: If (tn) is a bounded sequence and (sn) is a sequence such that limsn = 0 (Alternatively
of truncations of ?2 converges to ?2. 8.1d; CM Prove that lim n + 6 n2 ? 6. = 0. Fix ?>0. Choose N > 1 ? +6. If n>N then.
sn. = 0. b) Show that if lim n?? sn = 0 and sn > 0 for all n then lim n??. 1 sn. = ?. 2) Define {sn} by sn+1 = ?. 5 + sn for n ? 1 and s1 =.
(4) follows from (3) if we set tn = C for all n. Applying (4) with C = ?1 shows that limn(?tn) = ?t. Using this with (1)
case we define the sum of the series to be the limit of its partial sums. Proof. The series converges if and only if the sequence (Sn) of partial sums ...
(c) lim supsn is the biggest subsequential limit: 1; lim inf sn is the (a) We need to show that if (sn) is a convergent sequence of points in [a b]
9.12; 4pts Assume all sn =0 and that the limit L = lim\ sn+1 sn. . . exists. (a) Show that if L<1 then limsn =0. (b) Show that if L>1
We want to show that sn ? 0. We have ?
lim n?? sn = lim n?? sN+n ? lim n?? ans N = sN lim n?? an = 0 when a < 1 9 15 Show that limn?? an n! = 0 for all a ? R Put sn = an/n! and ?nd that sn+1/sn = a/(n + 1) tends to 0 as n ? ? Therefore by the previous exercise limsn = 0 (In other words n! grows faster than any exponential sequence an
Exercise 2 2Prove that lim n!1 3 n = 0 Exercise 2 3Prove that lim n!1 1 n2 = 0 Exercise 2 4Prove that lim n!1 ( 1)n n = 0 See Figure 2 3 Exercise 2 5Prove that lim n!1 1 n(n 1) = 0: It is good to understand examples when the de nition of converging to zero does not apply as in the following example Example 2 4Prove that the sequence s n= n+
lim n?? sn = s or by limsn = s or by sn ? s A sequence that does not converge is said to diverge Examples Which of the sequences given above converge and which diverge; give the limits of the convergent sequences THEOREM 1 If sn ? s and sn ? t then s = t That is the limit of a convergent sequence is unique Proof: Suppose s 6=t
(a) Show that if s n a for all but nitely many n then lims n a (b) Show that if s n b for all but nitely many n then lims n b (c) Conclude that if all but nitely many s n belong to [a;b] then lims n belongs to [a;b] 2 Section 9 9 2Suppose limx n = 3 limy n = 7 and all y n are nonzero Determine the following limits: (a) lim(x n + y n
(a)Show that if L < 1 then lims n = 0 Hint: Select a so that L < a < 1 and obtain N so that js n+1j< ajs njfor n N Then show js nj< an Njs Njfor n > N (b)Show that if L > 1 then limjsnj= +1 Hint: Apply (a) to the sequence t n = 1 sn; see Theorem 9 10 Proof (a) Since L < 1 we may choose a 2(L;1) Let " = a L Since js n+1 sn j!L
n;show that a+ bi= lim(a n+ b ni):That is a sequence fc n= a n+ b nigof complex numbers converges if and only if the sequence fa ngof the real parts converges and the sequence fb ngof the imaginary parts converges HINT: You need to show that given some hypotheses certain quantities are less than :Part (c) of Exercise 1 25 should be of help
n 2S (ii) lim 1 n = 0 so with (i) and (ii) and Theorem 11 9 we get that 0 2S 11 10 b Determine limsups n and liminf s n liminf s n = 0 and limsups n = 1 Note: There can be two ways to go about nding limsups n Either you use the de nition of limsup See page 60 Equation (1) limsups n = lim n!1 ( Supfs m for all m>ng): In which case you
Math 3150 Fall 2015 HW2 Solutions