121616949-math.256 - a n X b n In general the sequence of...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
242 Chapter 10 Sequences and Series So when | x | < 1 the geometric series converges to k/ (1 - x ). When, for example, k = 1 and x = 1 / 2: s n = 1 - (1 / 2) n +1 1 - 1 / 2 = 2 n +1 - 1 2 n = 2 - 1 2 n and X n =0 1 2 n = 1 1 - 1 / 2 = 2 . We began the chapter with the series X n =1 1 2 n , namely, the geometric series without the first term 1. Each partial sum of this series is 1 less than the corresponding partial sum for the geometric series, so of course the limit is also one less than the value of the geometric series, that is, X n =1 1 2 n = 1 . It is not hard to see that the following theorem follows from theorem 10.2 . THEOREM 10.17 Suppose that a n and b n are convergent sequences, and c is a constant. Then 1. X ca n is convergent and X ca n = c X a n 2. X ( a n + b n ) is convergent and X ( a n + b
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: a n + X b n . In general, the sequence of partial sums s n is harder to understand and analyze than the sequence of terms a n , and it is difficult to determine whether series converge and if so to what. Sometimes things are relatively simple, starting with the following. THEOREM 10.18 If ∑ a n converges then lim n →∞ a n = 0. Proof. Since ∑ a n converges, lim n →∞ s n = L and lim n →∞ s n-1 = L , because this really says the same thing but “renumbers” the terms. By theorem 10.2 , lim n →∞ ( s n-s n-1 ) = lim n →∞ s n-lim n →∞ s n-1 = L-L = 0 ....
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern