Proofs Flashcards
Prove that the limit of a convergent sequence is unique
Assume xn tends towards both L and M where L is not equal to M. Take epsilon to be |L-M|/2 > 0 and state definition of convergence for L and M. Take |L-M| and use triangle inequality to reach contradiction.
Prove that convergent sequences are bounded
Take epsilon to be 1 and state definition of convergence. For n bigger than or equal to n0, add and subtract L for modulus of xn and use triangle inequality to get that xn is smaller than |L|+1.
Take M to be the maximum of the modulus of all terms before xn0 and |L|+1. Then |xn| is less than or equal to M for all natural n.
Prove that xn+yn converges to L + M
State def. of convergence for xn and yn taking epsilon to be epsilon over 2. Use triangle inequality on |xn+yn-L-M| to reach less than epsilon.
Don’t forget to state that epsilon is arbitrary
Prove that xnyn converges to LM
xn is bounded by some K<0 by theorem. Take epsilon for x to be e/2K and epsilon for y to be e/2(|M|+1). Then add and subtract xnM to|xnyn-LM| and use triangle inequality to reach epsilon.
The sandwich principle
Let epsilon be arbitrary > 0. State definition of convergence for xn and zn and rearrange inequalities. Then use the bounds for yn and rearrange to get definition of convergence for yn for n bigger than max(n1,n2, N)
Monotone convergence sequence
Prove for ev. incr. and bounded above. Other case is similar. Define set S with xn where n is bigger than or equal to the N for which xn is increasing. Since S is non-empty and bounded, it has a sup, call it L, by completeness axiom. Claim that xn converges to L and let epsilon > 0 be arbitrary. Use definition of supremum and write out inequality to reach definition of convergence.
Use def. of sup for a specific n0 and then set n bigger or equal to n0
A sequence converges to L if and only if every subsequence converges to L.
Every sequence is a subsequence of itself so the implication from right to left is immediate.
Let xn converge to L as n goes to infinity and let k1<k2<… be a strictly increasing sequence of natural numbers defining a subsequence of (xn):
Let epsilon>0 be arbitrary. Use definition of convergence for some n0. Then for n bigger or equal to n0, kn is bigger than or equal to n big. eq to n0. Hence kn works in def of convergence. Write out implication and we have proven the statement.
The key is to show that kn is bigger than or equal to n0!
Comparison test
The second statement is the contrapositive of the first, so it’s
enough to prove the first statement. Show that partial sum sn for xn is bounded above and increasing. by MCT it converges.
Ratio test (1)
Assume n0=1. xn+1 is less than or equal to lambda times xn etc so 0 < xn < x1 lambda^(n-1) for all natural n. Since the series with lamda^(n-1) converges so does x1 lambda^(n-1) so by comparison test the series with xn converges.
Ratio test (2)
Assume n0=1.
xn+1 is bigger than or equal to xn etc > 0 so the sequence (xn): cannot converge to zero so the series diverges.
An absolutely convergent series converges.
- |xn| leq xn leq xn so 0 leq 2|xn| whose series converges. xn=xn+|xn|-|xn| so series with xn converges by algebraic properties.
f cts at c and g cts at f(c) implies g ◦ f is cts at c
State def for g of g and f using nu as delta for g and as epsilon for f. Put them together so that for delta epsilon.
