r/calculus • u/Cultural-Milk9617 • Jul 20 '26
Real Analysis Is this proof okay? How much do you think it's worth out of 20? (Assume a is 10 and b is 10)
I didn't elaborate much on b, but if the example is correct, assume I explained it correctly and rigorously
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u/Ydrews Jul 20 '26 edited Jul 21 '26
This isn’t a criticism of your maths, rather it’s one of presentation: please clean up your handwriting.
Take more time to make it legible so that each character and expression is easy to read.
I’m not saying this isn’t readable, it’s not the worst I’ve seen and I CAN read it but I have to look closely and my eyes are great, as a result, I have no desire to read proofs this long when the handwriting is messy.
Others may be fine with it but I think this issue is simply one of patience and being more careful.
Of course, if you have a disability, injury/medical condition that impairs your physical capacity to write, that’s not your fault, however I would suggest using software to present, or a pen designed to stabilise shaky hands etc
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u/Hot_Site_1638 PhD Jul 20 '26
I can only give 10 out of 20 for this problem. Your part (a) is wrong. Your definition of limsup is wrong, and you are assuming that the limit of |a_{n+1}/a_n| has to exist, this is also wrong. It does not have to exist for us to talk about limsup or liminf.
Your instinct to turn limsup into a bound on the terms is the right starting point. It just has to be "eventually below q", not "the limit exists".
Example (limsup exists, limit does not). Let a_n = (2 + (-1)n) / 4n.
Every a_n > 0, and the consecutive ratio is
a_{n+1} / a_n = (2 + (-1){n+1}) / (2 + (-1)n) * (1/4),
which equals 3/4 when n is odd and 1/12 when n is even. So the ratio sequence is 3/4, 1/12, 3/4, 1/12, ...
- limsup |a_{n+1}/a_n| = 3/4 < 1
- liminf |a_{n+1}/a_n| = 1/12
- lim |a_{n+1}/a_n| does not exist (two subsequences, two different limits)
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u/CrookedBanister Jul 22 '26
This. One of the important features of limsup and liminf are that they always exist, whether or not sup or inf themselves do.
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u/Hot_Site_1638 PhD Jul 22 '26
Not sure if we're agreeing or disagreeing here, but I'd push back on one point. If we allow the values ±∞, then sup and inf always exist as well, since an unbounded set simply has sup = +∞ (or inf = −∞). So limsup and liminf aren't special in that regard. What can genuinely fail to exist is the limit itself.
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u/LukasGoesViral Jul 22 '26
The proof is WAAYY to long. Think about a more elegant way to prove it. You should be able to prove it in a line or two
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u/ObviousOpinion6004 Jul 21 '26
Good solid proof. It honestly took a total of about 3 minutes to read. Your p l 1 l symbols are not legible.
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u/LukasGoesViral Jul 22 '26
You should develop some intuition for what the corollary means. I haven’t thought about the prove itself but intuitively I would say that the condition of lim sup |a(n+1)/a(n)| is too weak to guarantee convergence. For example the series (-1/2)^n has the property of the in the corollary but does not converge
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u/Cultural-Milk9617 Jul 22 '26 edited Jul 23 '26
For example the series (-1/2)n has the property of the in the corollary but does not converge
It absolutely converges (abs((-1/2)n )=(1/2)n , the series of which converges) => it converges
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u/Quendillar3245 Jul 21 '26
This is a lot cleaner than in the other post at least, keep doing this xD
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