319HW6Correction

# Square exercise 3 proof since a n n l one knows that

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square Exercise 3. Section 3.6, #10. Proof. Since ( a n / n ) L 0 one knows that lim ( a n ) =+ if and only if lim ( n ) =+ , which is the case, therefore lim ( a n )=+ . square Exercise 4. Section 3.7, #3a. Proof. For any natural number n greaterorequalslant 0 one has: 1 ( n +1) . ( n +2) = 1 n +1 - 1 n +2 . Therefore by adding these equalities for n =0 to n = N , one gets: n =0 n = N 1 ( n +1) . ( n +2) = n =0 n = N parenleftBig 1 n +1 - 1 n +2 parenrightBig = 1 1 - 1 N +2 1 when N + , so n =0 + 1 ( n +1) . ( n +2) =1 . square Exercise 5. Section 3.7, #6b. Proof. In class, using the Cauchy convergence criterion, I proved that if | x n | converges then so does x n . Here, since 0 lessorequalslant | cos n | n 2 lessorequalslant 1 n 2 , and we know that 1 n 2 converges, we deduce by the comparison theorem that | cos n | n 2 converges, and therefore cos n n 2 converges, by the above argument. square 1

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Exercise 6. Section 3.7, #8. Proof. Since a n is convergent, we know that necessarily ( a n ) 0 , therefore for n greaterorequalslant K for some large natural number K , we have that 0 < a n < 1 . Thus for n greaterorequalslant K one has 0 < a n 2 < a n , and then by the comparison theorem one deduces that a n 2 converges. square Exercise 7. Section 3.7, #11. Proof. Notice that b n 1/ n = ( a 1 + + a n ) converges to a n > 0 . Therefore by our comparison theorem, b n converges if and only if 1 n converges. Since it isn’t the case, we deduce that b n diverges. square 2
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