Convergent or Divergent: sum, n=1, infty, sin^2(x)/n^4

[MarkSA]

Hello,

If I have the series:
Summation from n=1 to infinity of: sin^2(x)/(n^4)

And so do the limit as n->infinity of: sin^2(x)/(n^4)

Why can the test for divergence not be used on this? Doesn't the limit of a sine function not exist since it is oscillating? Test for divergence states that if the limit does not equal 0 or does not exist, it is divergent.

 

[tkhunny]

Re: Convergent or Divergent

Are you sure you mean sin(x) and not sin(n)?

Anyway, \(\displaystyle 0 \leq sin^{2}(x) \leq 1\)

Do we care if it's oscillating if it's nicely bounded?

 

[MarkSA]

Sorry, yes it's n.

Doesn't it not approach a given value though, since it's always oscillating between 0 and 1?

I guess it doesn't matter by your question. Is the test for divergence on functions like sine/cosine then only valid if it's something like:

Summation from 1 to infinity of: sin^2(n)?

 

[soroban]

Re: Convergent or Divergent

Hello, Mark!

\(\displaystyle \sum^{\infty}_{n=1} \frac{\sin^2\!x}{n^4}\)

Use the Comparison Test.

We know that: . \(\displaystyle 0 \:\leq\: \sin^2\!x \:\leq \:1\)

\(\displaystyle \text{Then: }\;\frac{\sin^2\!x}{n^4} \:\leq \frac{1}{n^4}\)

\(\displaystyle \text{Hence: }\;\sum^{\infty}_{n-1}\frac{\sin^2\!x}{n^4} \;\leq \;\sum^{\infty}_{n=1}\frac{1}{n^4}\)

\(\displaystyle \text{The latter is a }p\text{-series with }p > 1\text{, hence it converges.}\)

\(\displaystyle \text{The former is less than a convergent series, and therefore converges.}\)