Monday, November 27, 2006

To Converge Or Not To Converge. Topic: Real Analysis/S&S.

Problem: (Stanford Math 51H, Cauchy Root Test) Suppose there exists a $ \lambda \in (0,1) $ and an integer $ N \ge 1 $ such that $ |a_n|^{\frac{1}{n}} \le \lambda $ for all $ n \ge N $. Prove that $ \displaystyle \sum_{n=1}^{\infty} a_n $ converges.

Solution: Well let's just discard $ \displaystyle \sum_{n=1}^{N-1} a_n $ because it is finite and obviously converges. It remains to show that $ \displaystyle \sum_{n=N}^{\infty} a_n $ converges.

But then we have $ |a_n|^{\frac{1}{n}} \le \lambda \Rightarrow |a_n| \le \lambda^n $. So

$ \displaystyle \sum_{n=N}^{\infty} a_n \le \sum_{n=N}^{\infty} |a_n| \le \sum_{n=N}^{\infty} \lambda^n $.

The last summation, however, is a geometric series with common ratio $ 0 < \lambda < 1 $, so it converges. Hence our sum does as well. QED.


Comment: The root test is, in effect, a comparison to a geometric series. The hypothesis is that we can bound $ |a_n| $ by a geometric series for all large $ n $, implying convergence.


Practice Problem: (Stanford Math 51H) Discuss the convergence of

$ \displaystyle \sum_{n=1}^{\infty} \frac{\sin{\frac{1}{n}}}{n} $.


  1. the inequality x ≥ sin x for positive x is quite easy to prove ;)

  2. "The word 'easy' should never be used in math."
    --gnay naux