convergence tests | Mathematics | Page 1

hidden vessel Jun 7, 2024, 4:34 AM

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Let $u_1$ be a real number, and $u_{n+1}=\frac{\sin(\frac{1}{n})}{e^{u_n}}.
Does $\sum_{n=1}^\infty u_n$ converges or diverges?

white dockBOT Jun 7, 2024, 4:34 AM

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TheVinkler
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fleet solsticeBOT Jun 7, 2024, 4:34 AM

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hidden vessel Let $u_1$ be a real number, and $u_{n+1}=\frac{\sin(\frac{1}{n})}{e^{u_n}}. Does...

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mental willow Jun 7, 2024, 4:57 AM

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for starters, does un converge to zero?

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also think of the ratio test

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@hidden vessel

hidden vessel Jun 7, 2024, 4:59 AM

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I think it does, it makes sense that it goes to 0 but I was unsuccessful to prove that

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sin(1/n) is aproximtly 1/n and so on but I couldn't prove it

mental willow Jun 7, 2024, 5:00 AM

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sin x ~ x when x -> 0, you can use this

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note that the product exp (un) * u(n+1) is bounded

hidden vessel Jun 7, 2024, 5:02 AM

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I know and I did , I got $u_{n+1}\approx \frac{1}{ne^{u_n}}$

white dockBOT Jun 7, 2024, 5:02 AM

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TheVinkler

mental willow Jun 7, 2024, 5:03 AM

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that's a start

hidden vessel Jun 7, 2024, 5:03 AM

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So what you are saying is something along the lines of

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$u_{n+1}\cdot ne^{u_n} \approx 1$

white dockBOT Jun 7, 2024, 5:03 AM

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TheVinkler

hidden vessel Jun 7, 2024, 5:04 AM

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oh you mean

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$u_{n+1}\cdot e^{u_n} \approx \frac{1}{n}$

white dockBOT Jun 7, 2024, 5:04 AM

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TheVinkler

mental willow Jun 7, 2024, 5:05 AM

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yes, so un must converge to zero because the right hand side tends to 0 whereas exp(un) is bounded below by 1

hidden vessel Jun 7, 2024, 5:07 AM

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mental willow yes, so un must converge to zero because the right hand side tends to 0 whereas ...

It's only bounded by 1 if un is positive, how do I justfy the negative?

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actally now I see

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u2 must be positive

mental willow Jun 7, 2024, 5:08 AM

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assume first u1 is positive

hidden vessel Jun 7, 2024, 5:08 AM

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ye ye

mental willow Jun 7, 2024, 5:08 AM

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doesn't really matter

hidden vessel Jun 7, 2024, 5:08 AM

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exp(u)>0

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sin(1/n)>0

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therefore u2>0

mental willow Jun 7, 2024, 5:09 AM

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or you can do induction if you wanna be formal about it

hidden vessel Jun 7, 2024, 5:10 AM

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Okay so un goes to 0, now just the ratio test and that's it?

mental willow Jun 7, 2024, 5:10 AM

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play around with it, maybe it diverges, who knows..

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if the un are positive eventually then you can also try comparing with another series

hidden vessel Jun 7, 2024, 5:11 AM

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I think ratio is the easiest

mental willow Jun 7, 2024, 5:12 AM

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go for it

hidden vessel Jun 7, 2024, 5:15 AM

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It isn't that easy despair

hidden vessel Jun 7, 2024, 5:20 AM

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mental willow go for it <:EZ:928304490468212806>

Ratio test isn't really working bcs I got $\lim_{n\to\infty}|\frac{1}{n\cdot u_n}|$

white dockBOT Jun 7, 2024, 5:20 AM

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TheVinkler

hidden vessel Jun 7, 2024, 5:21 AM

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and using the comperstion test I keep getting lower bounds converge but upper bounds diverge

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I thought about making a lower bound like this but it isn't very rigrous

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$\frac{\sin(\frac{1}{n})}{e^{u_n}}> \frac{\sin(\frac{1}{n})}{2}\approx \frac{1}{2n}$

white dockBOT Jun 7, 2024, 5:25 AM

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TheVinkler

hidden vessel Jun 7, 2024, 5:25 AM

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but I am not sure if this is even allowed

mental willow Jun 7, 2024, 5:32 AM

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yes because you know that un -> 0, so for large indices it must be true that exp (un) < 2

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@hidden vessel

hidden vessel Jun 7, 2024, 5:33 AM

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I know but the 2nd approx is what feels iffy

mental willow Jun 7, 2024, 5:33 AM

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it is correct

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if n i large enough then sin (1/n) ~ 1/n

hidden vessel Jun 7, 2024, 5:34 AM

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Do you think I have to justify it?

mental willow Jun 7, 2024, 5:34 AM

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it is already justified

mental willow Jun 7, 2024, 5:34 AM

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mental willow sin x ~ x when x -> 0, you can use this

it holds because of this

hidden vessel Jun 7, 2024, 5:35 AM

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technically speaking $\frac{\sin(\frac{1}{n})}{2}< \frac{1}{2n}$

white dockBOT Jun 7, 2024, 5:35 AM

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TheVinkler

mental willow Jun 7, 2024, 5:35 AM

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nono

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that is true

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but the two quantities are equivalent

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this is another form of comparison for convergence

mental willow Jun 7, 2024, 5:36 AM

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white dock **TheVinkler**

this formulation is sufficient

hidden vessel Jun 7, 2024, 5:36 AM

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ok

mental willow Jun 7, 2024, 5:36 AM

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you have shown the initial series is bounded below by something that is equivalent to a divergent series

hidden vessel Jun 7, 2024, 5:36 AM

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ye its between 2 harmonics therfore converges

mental willow Jun 7, 2024, 5:37 AM

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diverges!

hidden vessel Jun 7, 2024, 5:37 AM

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mb

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English isn't my first language

mental willow Jun 7, 2024, 5:37 AM

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however all of this assumes u1 >0

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does anything change if u1= 0 or negative?

hidden vessel Jun 7, 2024, 5:37 AM

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it doesn't really matter since u2>0

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so it doesnt change at all for those cases

mental willow Jun 7, 2024, 5:38 AM

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correct

hidden vessel Jun 7, 2024, 5:38 AM

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Wow, thanks a lot!

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I will keep this thread open until I finish writing my solution but thank you so much!

mental willow Jun 7, 2024, 5:39 AM

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hidden vessel ye its between 2 harmonics therfore converges

im confused about this bit, what does this mean

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it's not strictly speaking "between" anything, there is no sandwiching argument involved here

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are you familiar with the two main comparison tests?

hidden vessel Jun 7, 2024, 5:51 AM

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mental willow are you familiar with the two main comparison tests?

That's how I think abt it, but the thing that I am writing will just be abt how its bigger the 1/2n and that is sufficent for divergnce

hidden vessel Jun 7, 2024, 8:39 AM

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-close

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+close

#convergence tests