Question

Prove that the given series converges or diverges (below)

Answer 1

\[\sum_{n=1}^{\infty}nsin (\frac{ 1 }{ n })\]

Answer 2

Try the limit test?

Answer 3

Do you mean the limit comparison test?

Answer 4

Because if so, I have no idea what I would compare it to

Answer 5

Yeah \[a_n = n \sin (1/n)\]
\[\lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} n \sin (1/n)\]

Answer 6

let n = 1/x

Answer 7

Using divergence test

Answer 8

\[\lim_{x \rightarrow 0} 1/x \sin x\] our limit becomes as such then

Answer 9

Now you can apply L'Hospital's rule and notice it will diverge

Answer 10

Why did you pick 1/x?

Answer 11

.

Answer 12

And how did sinx get in the denominator

Answer 13

It's just a substitution and it's not.
\[\lim_{x \rightarrow 0} \frac{ 1 }{ x} sinx\]

Answer 14

I see now. Thank you!!!

Answer 15

Wait, why did you take the limit? Thats not one of the tests I was given

Answer 16

And why did you take the limit as x->0?

Answer 17

I used the divergence test

Answer 18

You learn it during integral test I think

Answer 19

Ok so let me just do it again

Answer 20

I know the limit comparison test, but thats different

Answer 21

I also know the nth term test, but thats for infinity

Answer 22

\[\large a_n = n \sin \left( \frac{ 1 }{ n } \right)\]
\[\large \lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} n \sin \left( \frac{ 1 }{ n } \right)\]
We know the series diverges if \[\large \lim_{n \rightarrow \infty} a_n \neq 0\]

We make a substitution \[\large n = \frac{ 1 }{ x }\] so as n goes to infinity we know x = 0

Answer 23

\[\large \lim_{x \rightarrow 0} \left( \frac{ 1 }{ x } \right) \sin(x) \implies \frac{ 0 }{ 0 } \implies \lim_{x \rightarrow 0} \cos(x) = \cos(0) = 1\]

Answer 24

So this is the nth term test?

Answer 25

Yes!

Answer 26

What about the times when sinx=0?

Answer 27

What's sin(0) ? :P

Answer 28

I'm very confused. The nth term test says that terms are getting larger (so it diverges) or smaller (so it converges)

Answer 29

@amistre64

Answer 30

It doesn't even prove convergence. If the terms are getting smaller, you have to do another test

Answer 31

@perl

Answer 32

\[n = \frac{1}{1/n}\]

Answer 33

Yes

Answer 34

Then I substituted x in for 1/n. Then I had no idea what to do

Answer 35

squeeze thrm .... you learn this sin(x)/x as x to zero very early on

might be useful here

Answer 36

Why do I care what the function is doing as x approaches 0? I'm supposed to prove that it converges or diverges

Answer 37

So we can change the limit

Answer 38

The only limit tests I was given involve infinity

Answer 39

Astro started it, good luck :)

Answer 40

I'm still confused :/

Answer 41

\[n = \frac{ 1 }{ x }\] so as n approaches infinity note that x = 0

Answer 42

x goes to 0

Answer 43

So our limit changes to \[\lim_{x \rightarrow 0} \frac{ 1 }{ x } \sin(x)\]

Answer 44

What do you mean as n approaches infinity x=0?

Answer 45

I meant x approaches inifinity

Answer 46

Do you mean the function approaches 0 as n-> infinity?

Answer 47

Yes

Answer 48

Ok

Answer 49

\[n -> \infty~~~x -> 0\]

Answer 50

Yes

Answer 51

I hope that makes sense now :)

Answer 52

So you substituted into the limit???

Answer 53

Yeah, we're just making a substitution for n.

Answer 54

I think you're just overthinking this

Answer 55

I probably am. Calculus is doing this to me. I think that I get it now though. So when you sub 0 in, it goes to infinity, so it diverges

Answer 56

No, it doesn't go to infinity, it's 1 so it diverges. \[\lim_{n \rightarrow \infty} a_n \neq 0\] for it to diverge.

Answer 57

if you plug 0 into 1/x sinx?

Answer 58

Please read over everything I've said

Answer 59

I'm confused again

Answer 60

What happens \[\lim_{x \rightarrow 0} \frac{ \sin(x) }{ x }\]

Answer 61

0/0->L'hopital->1
Lol of course I overthought it

Answer 62

Yes :)

Answer 63

I quit calculus

Answer 64

Naw, we all have these days

Answer 65

It happens sometimes when you are learning something new

Answer 66

The struggle is more important than the answer, so this way you learn better, it works out at the end :) at least I hope it does.

Answer 67

Yeah, I guess. I understood everything really well until we got to series. Thanks for helping me!!! I really appreciate it!

Answer 68

No worries, take care.

Answer 69

this is why people give up on something that is a beautiful as it is simple.

Answer 70

Wut lol