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ThePerfectHacker · #13 September 10th, 2007, 07:28 PM

The combinatorics question was answered by Soroban. Here is the solution to the series question. This question comes from my homework on Real Analysis last semester.

For any $\epsilon > 0$ rather consider $\frac{\epsilon}{2} > 0$ . Since the series is convergent by the Cauchy criterion it means $\left| \sum_{k=m}^n a_n \right| = \sum_{k=m}^n a_n < \frac{\epsilon}{2}$ for all $n\geq m\geq N$ . So if $m=N$ then it means $a_N+a_{N+1}+...+a_n < \frac{\epsilon}{2}$ . Since the sequence is non-increasing it means $(n-N)a_n = \underbrace{a_n+...+a_n}_{n-N} \leq a_{N}+...+a_n < \frac{\epsilon}{2}$ for $n\geq 2N$ . Also for $n\geq 2N$ we have $n = 2n - n \leq 2n -2N = 2(n-N)$ . This means $na_n \leq 2a_n(n-N) < \epsilon$ . Q.E.D.

Note: This problem gives us a elegant way to show the divergence of the harmonic series.