**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.

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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.

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

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

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

ReplyDelete"The word 'easy' should never be used in math."

ReplyDelete--gnay naux

haha nice xuan =)

ReplyDelete