Problem 4: critize the following proof find all mistakes(there may be more than one). A mistake includes but is not limited to a false statement; a statement that is not a logio consequence of its proceeding statement; a statement that is not usefull for or related to the final results. Prove that if {x2n} and {2n+1} converge to the same number L, then {xn} is also con- vergent to L. Proof By contradiction. Suppose {xn} is not convergent to L. Then it must have some subsequence {xp,} that is divergent. Now consider the common terms between {xp,} and {x2n}, they form a subsequence {Xqn} of both {xpn} and {r2n}. But on one hand, {xqn} is a subequence o {Xp, }, so it diverges. On the other hand, {rq,} is a subequence of {x2n}, so it converges to L. This is a contradiction.
Problem 4: critize the following proof find all mistakes(there may be more than one). A mistake includes but is not limited to a false statement; a statement that is not a logio consequence of its proceeding statement; a statement that is not usefull for or related to the final results. Prove that if {x2n} and {2n+1} converge to the same number L, then {xn} is also con- vergent to L. Proof By contradiction. Suppose {xn} is not convergent to L. Then it must have some subsequence {xp,} that is divergent. Now consider the common terms between {xp,} and {x2n}, they form a subsequence {Xqn} of both {xpn} and {r2n}. But on one hand, {xqn} is a subequence o {Xp, }, so it diverges. On the other hand, {rq,} is a subequence of {x2n}, so it converges to L. This is a contradiction.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Problem 4: critize the following proof
find all mistakes(there may be more than one).
A mistake includes but is not limited to a false statement; a statement that is not a logic
consequence of its proceeding statement; a statement that is not usefull for or related to the
final results.
Prove that if {x2n} and {2n+1} converge to the same number L, then {xn} is also con-
vergent to L.
Proof
By contradiction. Suppose {xn} is not convergent to L. Then it must have some subsequence
{xp,} that is divergent. Now consider the common terms between {xp,} and {x2n}, they form
a subsequence {Xqn} of both {xpn} and {r2n}. But on one hand, {xqn} is a subequence o
{Xp, }, so it diverges. On the other hand, {rq,} is a subequence of {x2n}, so it converges to
L. This is a contradiction.
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