Question

sum_{n=5}^{infty} 4sin \frac{ 2 }{ n } prove why it is divergent

Answer 1

\[\sum_{n=5}^{\infty} 4\sin \frac{ 2 }{ n }\]

Answer 2

Can it be proved by the limit comparison test?

Answer 3

Be careful, mate, since \(\sin(n)\to 0\), not \(1\).

Answer 4

(As \(n \to 0\))

Answer 5

Oops, totally was thinking in terms of \(\cos\frac2n\).

Answer 6

it should be divergent not convergent

Answer 7

Using a MacLaurin series, note that (1) it is alternating, and thus (2) it has an error of:
\[
R=\left(\frac{\frac{2}{n}}{3!}\right)^3
\]Or:
\[
R=\frac{1}{27n^3}
\]Noting this, we have that:
\[
\sin(x)\approx x
\]So, assuming a worst-case error:
\[
\sum_n\frac{2}{n}-\sum_n\frac{1}{27n^3}
\]But we know that the latter series converges, hence:
\[
\sum_n\frac{2}{n}-C
\]But, the former series diverges, hence:
\[
\sum_{n=5}^{\infty} 4\sin \frac{ 2 }{ n }
\]Also diverges.

Answer 8

Got it thank you

Answer 9

For more clarity, I should also add that:
\[
\sin(x)>x-\frac{x^3}{3!}
\]

Answer 10

Surely there's an easier way than invoking a small-angle approximation of \(\sin x\)... ?

Answer 11

I'm sure, I just forced a proof; too lazy to work nicely right now.

Answer 12

D'oh. This one is extremely easy. Compare against \(\frac4n\) in the limit like so:$$\lim_{n\to\infty}\frac{4\sin\frac1n}{\frac4n}=\lim_{n\to\infty}\frac{\sin\frac1n}{\frac1n}$$Let \(x=\frac1n\) and notice \(x\to0\) as \(n\to\infty\) so then we recognize \(0\lt\lim\limits_{x\to0}\frac{\sin x}{x}=1\lt\infty\), and since \(\sum\limits_n\frac4n=4\sum\limits_n\frac1n\) is a constant multiple of the harmonic series and thus diverges, it follows that \(4\sin\frac1n\), too, diverges.

Answer 13

@LolWolf's post is valuable advice but it's probably overkill in this particular example; the way we determine our guess, though, is recognizing that \(\sin x\approx x\) for small \(x\), which he validated using a Maclaurin series.