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Calculus BC: Series

The Comparison Test

Problems

Problems

Until the final section of this chapter, we will restrict our attention to series with a n≥ 0 . Thus the partial sums are increasing:

s 1s 2 ... s n ...    

If the series a n is to converge, there must be some B such that s nB for all n , or the s n will become arbitrarily large. Such a B is called an upper bound. The value to which the series converges is the least of all possible upper bounds. It turns out that whenever the sequence {s n} of partial sums has an upper bound, there exists a least upper bound, to which the series converges. This fact enables us to prove the comparison test, stated below.

For two series a n , b n , with a n, b n≥ 0 for all n , suppose there exists a number C > 0 such that

a nCb n    

for all n and that b n converges. Then a n converges and

a nC b n    

To prove this statement, it suffices to show that the number C b n is a bound for the partial sums a 1 + a 2 + ... + a n . Then the least upper bound of these partial sums must exist and is clearly less than or equal to C b n . Thus we need only note that


a 1 + ... + a n Cb 1 + ... + Cb n  
  = C(b 1 + ... + b n)  
  C b n  

A similar test enables us to show that certain series diverge. If a n and b n are again two series with a n, b n≥ 0 for all n , suppose that there exists C≥ 0 such that a nCb n for all n and that b n diverges. Then a n also diverges. The proof of this fact is similar to the previous proof--since the partial sums of the b n become arbitrarily large and

a 1 + ... + a nCb 1 + ... + Cb n = C(b 1 + ... + b n)    

the partial sums of the a n also become arbitrarily large.

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