Question

Does the series converge, absolutely converge, or diverge?

Answer 1

\[ \sum_{1}^{\infty} \frac{ 4+\sin(n) }{ \sqrt{n}}\]

Answer 2

n=1 ** for summation

Answer 3

you can compare this to the series 1 / sqrt(n)

Answer 4

we can make a direct comparison

Answer 5

in which case would be divergent because n^1/2, n < 1

Answer 6

?

Answer 7

@sikinder

Answer 8

$$
\Large \frac{4+\sin(n)}{\sqrt n } \geq \frac{4+(-1)}{\sqrt n }=\frac{3}{\sqrt n }

$$

Answer 9

but the series 3 / sqrt(n) diverges

Answer 10

$$
\Large{ \frac{4+\sin(n)}{\sqrt n } \geq \frac{4+(-1)}{\sqrt n }=\frac{3}{\sqrt n }
\\~\\
\sum\frac{4+\sin(n)}{\sqrt n } \geq \sum \frac{3}{\sqrt n }

}$$

Answer 11

By the comparison test, because the original series is larger than 3/n^1/2, which diverges, the original diverges as well ?

Answer 12

$$
\Large{ \frac{4+\sin(n)}{\sqrt n } \geq \frac{4+(-1)}{\sqrt n }=\frac{3}{\sqrt n }
\\~\\\Rightarrow ~ \sum_{n=1}^\infty\frac{4+\sin(n)}{\sqrt n } \geq \sum_{n=1}^\infty \frac{3}{\sqrt n } = \infty
\\~\\\Rightarrow ~ \sum_{n=1}^\infty\frac{4+\sin(n)}{\sqrt n } \geq \infty

}$$

Answer 13

thanks

Answer 14

correct, and do you see why sin(n) >= -1, because of the sine graph

Answer 15

because sin(x) >= -1

Answer 16

Yes, the sine function is bounded between -1 and 1

Answer 17

this argument is not super rigorous, but i don't want to lose you in the details of a formal proof.

Answer 18

or you can add details to make it more rigorous. but i think you get the idea

Answer 19

yeah, it makes sense. I obviously need to review my inequalities however. Thanks again :)

Answer 20

i would have to cite theorems that i used

Answer 21

sure :)

Answer 22

i feel the urge to add a few details
$$
\large{ \frac{4+\sin(n)}{\sqrt n } \geq \frac{4+(-1)}{\sqrt n }=\frac{3}{\sqrt n }
\\~\\\Rightarrow ~ \sum_{n=1}^\infty\frac{4+\sin(n)}{\sqrt n }
\geq \sum_{n=1}^\infty \frac{3}{\sqrt n } = 3\cdot \sum_{n=1}^\infty \frac{1}{ n^{1/2} }=3\cdot \infty
\\~\\\rm because ~ p ~ series \sum_{n=1}^\infty \frac{1}{ n^{p} }~~diverges ~to~ \infty ~if~ p \leq 1
\\~\\\Rightarrow ~ \sum_{n=1}^\infty\frac{4+\sin(n)}{\sqrt n } \geq \infty

}$$

Answer 23

Ops no need for all these complications.
Simply you split your signa into two Then study each

Answer 24

The first would be p-series with p=0.5 <1 it means diverge.
Then the whole series diverge regardless of your second series.
I hope it is helpful:)