## 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.

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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

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

1. 2. 3. 