Using the Ratio Test, I have to find whether $$ \sum_{n=1}^\infty \frac{\cos(n\pi/3)}{n!} $$ converges or diverges. The back of the book says that the sum is absolutely convergent.

My work:

$a_n = \dfrac{\cos(n\pi/3)}{n!}$, $a_{n+1} = \dfrac{\cos((n+1)\pi/3)}{(n+1)!}$

\begin{align} &\lim_{n\rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right| \\[6pt] \implies&\lim_{n\rightarrow \infty} \left|\frac{\dfrac{\cos((n+1)\pi/3)}{(n+1)!}}{\dfrac{\cos(n\pi/3)}{n!}}\right| \\[12pt] \implies&\lim_{n\rightarrow \infty} \left|\frac{\cos((n+1)\pi/3) \cdot n!}{\cos(n\pi/3)\cdot(n+1)!}\right| \\[6pt] \implies&\lim_{n\rightarrow \infty} \left|\frac{\cos((n+1)\pi/3)}{\cos(n\pi/3)\cdot(n+1)}\right| \\ \end{align}

Now this is where I am stuck. I don't know how to find the limit for the $\cos$ terms. I tried using the identity $\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$ but it didn't yield anything useful (maybe, I should have tried harder?). I tried looking at this question but it didn't help much.

Any hints would be appreciated.

Thanks for your time!

  • 1
    $\begingroup$ it's simpler than that. how can you bound cosine from above ? $\endgroup$ – Glougloubarbaki Apr 19 '15 at 23:23
  • $\begingroup$ It's bounded by -1 and 1 but since it's absolute value, it would just be 1. $\endgroup$ – Jeel Shah Apr 19 '15 at 23:24
  • $\begingroup$ hint : for $n>3$ $$|\frac{cos(\frac{n\pi}{3})}{n!}|<\frac{1}{n^2}\\so\\\sum |\frac{cos(\frac{n\pi}{3})}{n!}|<\sum \frac{1}{n^2}=\frac{\pi^2}{6}$$hint : for $n>3$ $$\frac{}{} $$ $\endgroup$ – Khosrotash Apr 19 '15 at 23:25
  • $\begingroup$ Ratio test is problematic when the series isn't absolutely decreasing - that is, where $|a_n|$ isn't a decreasing sequence. $\endgroup$ – Thomas Andrews Apr 19 '15 at 23:27
  • $\begingroup$ As a general rule, factorials are huge. Cosines are tiny, never more than $1$. $\endgroup$ – davidlowryduda Apr 20 '15 at 0:09

Use comparison. $${|cos(n\pi x/3)|\over n!} \le {1\over n!}.$$

  • $\begingroup$ I can't use a comparison test unfortunately :( $\endgroup$ – Jeel Shah Apr 19 '15 at 23:25
  • 3
    $\begingroup$ Really? The comparison test is way simpler than the ratio test. $\endgroup$ – Thomas Andrews Apr 19 '15 at 23:28
  • 2
    $\begingroup$ Combining the two is the correct tool to do this job. $\endgroup$ – ncmathsadist Apr 19 '15 at 23:29
  • $\begingroup$ @ThomasAndrews This is kind of a stupid reason but it's in the "ratio test" section so I figure might as well practice ratio test with problems which are easier done with other methods but if this was on an exam then I would be more than happy to run with the comparison test. :D $\endgroup$ – Jeel Shah Apr 19 '15 at 23:34
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    $\begingroup$ @JeelShah But you can use the comparision test and then the ratio test to show, that $\sum (1/n!)$ is convergent, can't you? The comparision test used? Used! :-) $\endgroup$ – Kola B. Apr 19 '15 at 23:43

Why so complicated? It is dominated by $\sum (1/n!)$, which is obviously convergent.

  • $\begingroup$ Use the ratio test is part of hte problem $\endgroup$ – Jwan622 Apr 9 at 2:51

the sum (1*) is: $$ \Re \left(\sum_{n=1}^\infty \frac{ (e^{\frac{i\pi}3})^n}{n!}\right) = \Re (e^{e^\frac{i\pi}3}) = \sqrt{e}\cos(\frac{\sqrt{3}}2) $$

1* the complex series gives $$ |\frac{a_{n+1}}{a_n}| = \frac{|e^{\frac{i\pi}3}|}{n+1} $$ hence is seen to be absolutely convergent


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