/sci/ - Science & Math » Thread #13273886

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Anonymous Sun 13 Jun 17:54:28 2021 No.13273886 View Reply Original Report

Quoted By: >>13273915

Can someone that knows Mathematical Analysis or has a Wolfram Alpha account help me with this problem ? I can't use differentials, integrals or series to solve it, need to rely on the basic stuff from sequences like the squeeze theorehm, Stoltz etc.

Anonymous

Anonymous Sun 13 Jun 2021 18:06:39 No.13273915 Report

Quoted By: >>13273924

>>13273886
So it's a paywalled AI that can be exploited by any student to cheat at math homework?
Are you going to fork over cold hard cash?
Don't they teach L'Hopital's rule anymore?

Anonymous

Anonymous Sun 13 Jun 2021 18:09:58 No.13273924 Report

Quoted By:

>>13273915
Don't cheat.
No.
Yes they do, but they like saying without using L'Hopitals rule on tests.

Anonymous

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Anonymous Sun 13 Jun 2021 19:04:13 No.13274144 Report

Quoted By: >>13274372

To show in your pic is
$\lim_{n\to \infty} n \left( \left(\frac{n+1}{n}\right)^n - {\mathrm e} \right) = -\frac{{\mathrm e}}{2}$

I feel the way in which this is posed is weird and it would be simpler to think about after inverting both sides and flipping the signs.
What is really at stake here it to show that

$n \left( {\mathrm e}^{-x} \left(1+\frac{x}{n}\right)^n - 1 \right) = -\frac{x^2}{2} + O(\frac{1}{n})$

or

${\mathrm e}^{-x} \left(1+\frac{x}{n}\right)^n = 1 - \frac{x^2}{2n} - \frac{x^3}{3n^2} + O(\frac{1}{n^3})$

And then your theorem is a lemma at x=1.
It's simple to show via Taylor expansion to the third term, but I haven't thought about solving it via squeeze theorem, but that's what behind this

Anonymous

Anonymous Sun 13 Jun 2021 20:14:38 No.13274372 Report

Quoted By:

>>13274144
Thank you, at least i know what to use after the guidance

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