barcode in ssrs report CAUCHY S CONVERGENCE CRITERION 2.24. Prove Cauchy s convergence criterion as stated on Page 25. in .NET

Painting QR in .NET CAUCHY S CONVERGENCE CRITERION 2.24. Prove Cauchy s convergence criterion as stated on Page 25.

CAUCHY S CONVERGENCE CRITERION 2.24. Prove Cauchy s convergence criterion as stated on Page 25.
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Necessity. Suppose the sequence fun g converges to l. Then given any  > 0, we can nd N such that jup lj < =2 for all p > N Then for both p > N and q > N, we have jup uq j j up l l uq j @ jup lj jl uq j < =2 =2  Su ciency. Suppose jup uq j <  for all p; q > N and any  > 0. Then all the numbers uN ; uN 1 ; . . . lie in a nite interval, i.e., the set is bounded and in nite. Hence, by the Bolzano Weierstrass theorem there is at least one limit point, say a. If a is the only limit point, we have the desired proof and lim un a. Suppose there are two distinct limit points, say a and b, and suppose b > a (see Fig. 2-1). By de nition of limit points, we have jup aj < b a =3 for infinnitely many values of p juq bj < b a =3 for infinitely many values of q Then since b a b uq uq up up a , we have jb aj b a @ jb uq j jup uq j jup aj 3 1 2
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b_a 3 a b_a 3 b
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juq lj < =2 for all q > N
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Fig. 2-1
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Using (1) and (2) in (3), we see that jup uq j > b a =3 for in nitely many values of p and q, thus contradicting the hypothesis that jup uq j <  for p; q > N and any  > 0. Hence, there is only one limit point and the theorem is proved.
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INFINITE SERIES 2.25. Prove that the in nite series (sometimes called the geometric series) a ar ar2 (a) converges to a= 1 r if jrj < 1,
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Let Then Subtract,
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(b) diverges if jrj A 1.
ar ar2 arn 1 arn arn
Sn a ar ar2 arn 1 rSn 1 r Sn a
SEQUENCES
[CHAP. 2
Sn If jrj < 1; lim Sn lim
n!1 n!1
a 1 rn 1 r
a 1 rn a by Problem 7: n!1 1 r 1 r
(b) If jrj > 1, lim Sn does not exist (see Problem 44).
2.26. Prove that if a series converges, its nth term must necessarily approach zero.
Since Sn u1 u2 un , Sn 1 u1 u2 un 1 we have un Sn Sn 1 . If the series converges to S, then
lim un lim Sn Sn 1 lim Sn lim Sn 1 S S 0
n!1 n!1 n!1
2.27. Prove that the series 1 1 1 1 1 1
1 X 1 n 1 diverges. n 1
Method 1: lim 1 n 6 0, in fact it doesn t exist. Then by Problem 2.26 the series cannot converge, i.e., it diverges.
Method 2: The sequence of partial sums is 1; 1 1; 1 1 1; 1 1 1 1; . . . i.e., 1; 0; 1; 0; 1; 0; 1; . . . . Since this sequence has no limit, the series diverges.
MISCELLANEOUS PROBLEMS 2.28. If lim un l, prove that lim
n!1 n!1
u 1 u2 un l. n
Let un vn l.
v1 v2 vn 0 if lim vn 0. Now n!1 n v1 v2 vn v1 v2 vP vP 1 vp 2 vn n n n We must show that lim
so that
  v 1 v 2 v n    @ jv1 v2 vP j jvP 1 j jvP 2 j jvn j   n n n Then
Since lim vn 0, we can choose P so that jvn j < =2 for n > P.
jvP 1 j jvP 2 j jvn j =2 =2 =2 n P =2  < < n n n 2 After choosing P we can choose N so that for n > N > P, jv1 v2 vP j  < n 2 Then using (2) and (3), (1) becomes   v 1 v 2 v n     <    2 2 n thus proving the required result.
for n > N
2.29. Prove that lim 1 n n2 1=n 1.
Let 1 n n2 1=n 1 un where un A 0.
Now by the binomial theorem,
CHAP. 2]
SEQUENCES
1 n n2 1 un n 1 nun
n n 1 2 n n 1 n 2 3 un un un n 2! 3!
Then 1 n n2 > 1 Hence, lim u3 0 and n
n n 1 n 2 3 6 n2 n un or 0 < u3 < : n 3! n n 1 n 2 lim un 0: Thus lim 1 n n2 1=n lim 1 un 1:
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