determine weather the sum of (3sin(n))/n^(5/3) is convergent or divergent from 0 to infinity

Question

hero · Answer

anonymous · Answer

i would imagine it converges

anonymous · Answer

since n^a, a is bigger than 1

anonymous · Answer

it is convergent

anonymous · Answer

well 
$$\frac{5}{3}>1$$[

anonymous · Answer

and so 
$$\sum\frac{1}{n^{\frac{5}{3}}}$$ converges

hero · Answer

I don't reallly know much about converge and diverge. I just found series expansion on wolfram alpha. I see that it is not reliable

anonymous · Answer

and 
$$\sin(n)\leq 1$$

anonymous · Answer

wow i can't even read that. but we can use wolfram to check

anonymous · Answer

hero · Answer

So that part that is posted on wolfram alpha that I posted, what do you call that?

anonymous · Answer

so what is the first step? can i do the comparison test?

hero · Answer

I will study these more so that I know the steps.

anonymous · Answer

yes you can compare it to a "p" series ( i think that is what it is called)

anonymous · Answer

$$\sum\frac{1}{n^p}$$ converges if
$$p>1$$

amistre64 · Answer

$$ \frac{3sin(n)}{n^{5/3}}$$
sin(n) is alternating

anonymous · Answer

well not exactly

amistre64 · Answer

almost got that right lol

anonymous · Answer

lol

amistre64 · Answer

cos(n) alternated ;)

amistre64 · Answer

...pi*n that is , ill get my head on straight eventually

anonymous · Answer

but in any case it  is true that 
$$\sin(n)\leq 1$$so 
$$\frac{\sin(n)}{n^{\frac{5}{3}}}\leq \frac{1}{n^{\frac{5}{3}}}$$

anonymous · Answer

one on the right converges so one on the left does as well.

anonymous · Answer

yeah that pi ...