Question

Curious if this is the right way
looking for the convergence of
\[\sum \sin (\frac{4n^2+5}{n^4+6})\]

Answer 1

this is what i did
\[\frac{4n^2+5}{n^4+6}<\frac{4}{n^4}\]
then
\[\sin (\frac{4n^2+5}{n^4+6})<\sin (\frac{4}{n^4})\]
the series on the right converges using p series test

Answer 2

I'm using the fact that sin is continuous therefore
\[\sum \sin (4/n^4)=sin(\sum 4/n^4)\]

Answer 3

Your conclusion is correct, but reasoning isn't..
sin(x) is not a strictly increasing function

Answer 4

oh one little mistake the right hand of the inequality should have been 4/n^2

Answer 5

you may use this fact for \(x\gt 0\) instead :
\[\sin(x) \lt x\]

Answer 6

hmm i see, just look for the convergence of what's inside using that fact

Answer 7

and conclude the convergent by comparison test

Answer 8

gotcha :)
thanks

Answer 9

Yes\[\sin \frac{4n^2+5}{n^4+6} \lt\frac{4n^2+5}{n^4+6}<\frac{4n^2+5}{n^4} \]

need to massage further..

Answer 10

that would be less than 4/n^2 yes?

Answer 11

which converges by p series

Answer 12

i don't think thats obvious..

Answer 13

how is
\[\dfrac{4n^2+5}{n^4}\lt \dfrac{4}{n^2}\]

?

Answer 14

yeah it really is not hehe.
we use limit comparison test

Answer 15

thats a good idea !!

Answer 16

but what sequence are we gonna use! i still have some subtlety with this test
4/n^2

Answer 17

lim 4/n^2/ (4n^2+6)/n^4
how do we decide the right sequence to do the limit comparison test

Answer 18

the book just mentions that the denominator behaves like a know sequence
then they conclude the convergence with the limit if it is finite or not!
I'm using the same technique, but it is not a guarantee for some cases

Answer 19

known*

Answer 20

gotta go, thanks for the help :)

Answer 21

that should work nicely right
\[a_n:=\dfrac{4n^2+5}{n^4} \\~\\b_n:= \dfrac{1}{n^2}\]

\[\lim\limits_{n\to\infty}~\dfrac{a_n}{b_n}=\lim\limits_{n\to\infty}~\dfrac{4n^2+5}{n^2}=4\]

Since \(4\) is finite and the series \(\sum\limits_{n=1}^{\infty}b_n\) converges, the series \(\sum\limits_{n=1}^{\infty}a_n\) also converges.