Calc6.57

From Example Problems
Jump to: navigation, search

Discuss the convergence or divergence of the series \sum _{{k=1}}^{{\infty }}\sin {\frac  {2k\pi }{n}}{\frac  {1}{\ln(\ln(\ln k))}} for any integer n\geq 1.

Let us first look at the sequence of partial sums

\{s_{n}\}=\left\{\sum _{{k=1}}^{{n}}\sin {\frac  {2k\pi }{n}}\right\}

This sequence is bounded, though by a different number for each n. If n is even, the first n values of k will be split up with two values of k that give a 0 for the sine function, k={\frac  {n}{2}} and k=n. Of the remaining n-2 values, the first half, a total of {\frac  {n}{2}}-1, will give a positive value for sine and the second half will cancel those all out. Thus, an easy to calculate bound (we just need to show that there is one, not find the least upper bound) is

0\leq \{s_{n}\}\leq {\frac  {n}{2}}-1

since sine can be at most 1. If n is odd, this is only slightly different. Now, the nth value of k will give a 0 sine, where as the first {\frac  {n-1}{2}} will give a positive sine and the next {\frac  {n-1}{2}} will give exactly opposite negative sines, canceling out the first. Thus, for any n, the sequence of partial sums is bounded by

0\leq \{s_{n}\}\leq {\frac  {n-1}{2}}.

Now, since

{\frac  {1}{\ln(\ln(\ln k))}}

goes to 0 as k\rightarrow \infty , by Dirichlet's test, this series converges.


Main Page : Calculus