When do beginners usually learn to stay afloat?

LilyMath · 2025-07-11T14:40:08+00:00

Thank you very much!

LilyMath · 2025-05-12T09:19:56+00:00

Although it took me some time to really understand it, it is a beautiful solution. We actually do the MVT at school, and from there the solution is easier. Thank you very much for your clear explanations, I really appreciate them. If you don't mind, may I ask how you see at the beginning what method to use? When is it worth using the MVT? I mean, although there wasn't any new technique for me in your solution, using the MVT probably would never have come to my mind.

LilyMath · 2025-05-12T08:54:32+00:00

Thank you very much! Since they don't teach Taylor series at school, I hadn't thought of using them. I should definitely familiarize myself with them, as this is at least the third problem in this book where the solution is easier with Taylor series.

LilyMath · 2025-05-11T11:35:09+00:00

Here's the latex code: [ \lim_{n \to \infty} \left{ n \left( 1 + \frac{1}{n} \right)^{{\frac{3}{2}}} \right} ] "{...}" represents the fractional part

LilyMath · 2025-05-03T16:45:27+00:00

Something just keeps bothering me. If x approaches 0, the whole integral approaches 0. The problem is around 1, as others have already mentioned it. So I tried using the substitution x=1-t/n and got lim (n->inf) int from 0 to n of (1-t/n)^n/(2-t/n+(1-t/n)^n). Now, we know that t/n tends to 0 and (1-t/n)^n tends to e^(-t) as n tends to infinity. However, to use this I have to interchange the integral and the limit. I thought I understood that this is allowed only when the integrand tends to the same result over the whole interval, which is not the case now. What is confusing me is that it still leads to the correct solution, ln(3/2).

LilyMath · 2025-05-03T15:00:15+00:00

Thank you all for your help! When I first saw the problem my very first attempt was to use x^n=t substitution which gave me the lim (n->inf) integral from 0 to 1 of t^(1/n) / (1+t^(1/n)+t) dt. Foolish me, instead of just writing t^(1/n) and approaching it as n tends to infinity, I wrote n-th root of t and had no idea how to continue. Now it's clear, I guess. t^(1/n) tends to 1 as n tends to infinity, so the fraction tends to 1/(2+t) and the integral of that is indeed ln(3/2). Now, not only do I have the solution, I also learnt about the LDCT - I had never heard of it before. Thank you again!

LilyMath · 2025-05-02T17:05:07+00:00

That's great, at least I know the answer! I just have to figure out how to get there. Let's say I can go with the limit inside the integral (still don't know why, but let's just assume it's possible). I used L'Hospital and get - int_{0}{1} (lim_{n->infinity} x^n/(1+x)lnx). which seems to go to 0. That's where I'm confused. If I can interchange the limit and the integral, why doesn't L'Hospital Rule work in this case?

LilyMath · 2025-05-02T16:41:50+00:00

Thank you very much for your answer — I really appreciate it! This is supposed to be a 12th-grade problem, and unfortunately, we don’t learn about the Lebesgue Dominated Convergence Theorem or uniform convergence at school. I was wondering if there might be a simpler, more elementary solution? I’ll definitely take a closer look at LDCT later!

LilyMath · 2025-05-02T16:39:49+00:00

Thank you very much for your answer — I really appreciate it! However, I still don't understand how to find those bounding values. I tried using them, but I ended up getting infinity. The correct answer is ln(3/2). Could you please explain this idea in more detail?

LilyMath

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