How can I go about showing that the series $\sum_{n=1}^\infty \frac{\cos(nx)}{n^x}$converges for $x>1$ ?
Asked
Active
Viewed 183 times
1
-
1What is the upper bound for $| \cos x |$? – Rellek Sep 02 '16 at 21:22
2 Answers
5
We have that $\frac{|\cos(nx)|}{n^x}\leq \frac{1}{n^x}$ and $\sum_{n=1}^{\infty}\frac{1}{n^x}<+\infty$ for $x>1$. So by Comparison Test, $\sum_{n=1}^\infty \frac{\cos (nx)}{n^x}$ is absolutely convergent and therefore it is also convergent.
P.S. Note that the series is conditionally convergent for $x=1$: conditional convergence of $\sum_{n=2}^{\infty} \frac{\cos(n)}{n}$
-
1ahh ok I see, so it converges automatically since 1/n^x converges for x>1 ? – dfge34 Sep 02 '16 at 21:30
-
@dfge34 use comparison test and the fact that absolute convergence implies convergence. – Robert Z Sep 02 '16 at 21:35
1
Hint: Use Dirichlet's test .
Show that for every positive $N$ there such $M$ number that $\left| \sum _{ n=1 }^{ N } \cos (nx) \right| \le M$ is bounded and $\frac { 1 }{ n^{ x } } $ monotone decreasing
haqnatural
- 21,578
-
1$\sum_{n=1}^{\infty}\cos(nx)$ is not bounded, it doesn't converge. There exists a number $B$, such that for any $N$, $\left|\sum_{n=1}^{N}\cos(nx)\right|\le B$. – Mark Viola Sep 02 '16 at 22:17
-
1You're saing that it's not bounded in the first sentence, and then saying it's bounded in the second. – xyzzyz Sep 02 '16 at 22:31