Advanced Calculus please help!!! 0. Let trn a bounded sequence and let z lim(zm)

Question

Advanced Calculus please help!!!

0. Let trn a bounded sequence and let z lim(zm). Prove that there exists E 0 and an integer N such that for every n N, zn r E. 1. Let frn and tynt two bounded sequences a. Show that lim (rn yn) S lim(an) lim(yn) (Hint: prove that if z lim(an) lim (yn), then z cannot be a cluster point of the sequence tzn yr). You may need the result of question 0. at some point) {rn), tynt where the inequality in a) is an equality b. Find an example of sequences c. Find an example of sequences tznh, tynt where the inequality in a is strict 2. Let tan) a bounded sequence. Denote M the least upper bound of the set S n E Nt. a. Prove that lim (rn) S M. b. Find an example of sequence zn such that lim (zn): M c. Find an example of sequence zm such that lim(zn) M.

Explanation / Answer

I'll just do the exercises about power series, there are far too many questions for 1500 pts...

I suggest you post what I didn't do in another thread.

1)

(a) an = c^(sqrt(n)) with c>0

Using root test we get an^(1/n) = c^(sqrt(n)/n) which has limit 1 at infinity

(since sqrt(n)/n converges to 0 and c>0)

So R = 1

(b) Here using root test again we see that an^(1/n) = (1+1/n)^n which converges to e

So R = 1/e

(c) The first part of the series is (2x)^(2n) and the second part is (3x)^(2n+1)

By the geometic series test, we need |2x|<1 for the first and |3x|<1 for the second

So |x| < min {1/2,1/3} = 1/3

So R = 1/3

(d) Using ratio test this time :

|a(n+1)/a(n)| = (n+1)! / n! * n^n / (n+1)^(n+1) = [n/(n+1)]^n = [1-1/n]^n which has limit 1/e

So R = e

(e) Using root test a(n)^(1/n) = 1/sqrt(n) which converges to zero.

So R = +infinity (convergence everywhere)

(f) Using root test a(n)^(1/n) = 1/n^(sqrt(n)/n) which converges to 1 since sqrt(n)/n converges to 0.

So R=1

4) Here you include the x term in the test they are different from x^n

(a) Using ratio test with a(n) = x^(n^2)/2^n then : |a(n+1)/a(n)| = |x^(n+1)^2 / x^(n^2) * 2^n / 2^(n+1)|

But x^(n+1)^2 / x^(n^2) =x^(2n+1)

So : lim(n->+infinity) |a(n+1)/a(n)| = +infinity if x>1 and lim(n->+infinity)| a(n+1)/a(n) |= 0 if x<1

(Note : at x=+-1 the series converges by geometric series test since 1/2 < 1, however it's not required)

So the series converges for |x|<=1

So R = 1

(b) Using ratio test : |a(n+1)/a(n)| = (n+1) x^((n+1)!-n!) = (n+1) x^( n * n!)

So we see that for |x|>=1 the limit will be infinity and for |x| <1 the limit is clearly 0 < 1 (converges)

So R=1

(c) Using ratio test : |a(n+1)/a(n)| = (n+1)^(n+1) / n^n x^(2n+1)

= ((n+1)/n)^n * (n+1) x^(2n+1)

The left part ((n+1)/n)^n converges to e, however (n+1)x^(2n+1) can't have limit < 1 unless |x| < 1

(for which limit is zero)

So R=1 (again)

(d) Using ratio test : |a(n+1)/a(n)| = r^(n+1)^2-n^2 = r^(2n+1) which has limit 0 has 0<r<1

So R=+infinity (converges everywhere)

(e) Same reasonning than in (c) : |a(n+1)/a(n)|= r x^(2n+1) , which converges to 0 if |x|<1 and +infinity if |x|>1

So R=1

(f) |a(n+1)/a(n)| = n x^(2n+1) ....again same converges to 0 < 1 if and only if |x|<1

So R=1

2)

Writing this as : sum(n>=0) sum(k=0..n) a(k)x^k x^(n-k) we recognize a cauchy product which is :

[ sum(n>=0) a(n) x^n ] * [ sum(n>=0) x^n ] = sum(n>=0) a(n) x^n/(1-x) (as soon as |x|<1)

(a) Here a(n) = 1/n! so this is 1/(1-x) * sum(n>=0) x^n / n! = e^x/(1-x)

(b) Here a(n) = (-1)^(n+1)/n for n>=1 so the sum is 1/(1-x) * sum(n>=1) (-1)^(n+1) / n x^n = ln(1+x)/(1-x)

(c) Here a(n) = n for n>=0 so the sum is :

1/(1-x) sum(n>=1) nx^n = x/(1-x) * sum(n>=1) nx^(n-1) = x/(1-x) * d/dx [ 1/ (1-x) ] = x/(1-x)^3

5) We already proved this in 4) for (c) , (e) and (f) if you look carefully ...

Using ratio test we get with a(n)x^(n^2) we get : | a(n+1) / a(n) |* x^(2n+1), since |a(n+1)/a(n)| converges to 1/R by assumption then we need x<1 to have a finite limit (0 <1 , so converges), otherwise x^(2n+1) will diverges to +infinity ( > 1 : diverges)

So R=1

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