a very cool problem with perfect squares.

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[@miraj2264]

Funnily enough, at 5:15 you noted that the difference between two squares (m+1)^2 and m^2 is 2m+1. This can be leveraged later to avoid going through the partial fraction decomp: (2n+1) / [(n^2)(n+1)^2] = [(n+1)^2 - n^2] / [(n^2)(n+1)^2] = [1/(n^2)] - [1/(n+1)^2] If you wanted to go the partial fraction route, it's probably easier to work with the equation at 12:09 given that n and n+1 both appear in 3 out of 4 terms. Simply plug in n = 0 ==> 1 = B and n = -1 ==> -1 = D. I believe this is called the Heavyside Coverup method albeit rearranged a bit. At this point, the contribution from the B/D terms alone makes the RHS equals the LHS. Thus the contribution from the A/C terms must be 0. Since A/C terms are linearly indep, A=C=0.

[@Jack_Callcott_AU]

Clever thinking.

[@koenth2359]

Tried from the thumbnail, but it says b_n

[@wesleydeng71]

10:37 Notice that(n+1)^2 - n^2 = 2n+1, saving your 5 minutes.

[@dieterjagutis6226]

(2n+1)/(n^2*(n+1)^2 (1) and we know 2n+1 = (n+1)^2-n^2 (2) so put (2) in (1) we give ((n+1)^2 - n^2) /(n^2 * (n+1)^2) = 1/n^2 - 1/(n+1)^2 without compute constants A, B, C and D Nice explanation of the problem. Thanks.

[@DrR0BERT]

That is very simple.

[@cultist_pixel]

Can i suggest a lil math thingy i once saw in some russian math contest? It goes like this: there is a polynomial Pn(x)=x^n + a(n-1)*x^(n-1) + ... + a1*x + 1. Polynomial has n roots, all a(i) are positive/non-negative. Prove that Pn(2023)>=2024^n.

[@khayalaliyev3519]

It is easy

[@tomkerruish2982]

Are we assuming that the coefficients are all (positive) integers?

[@khayalaliyev3519]

If you want I can send you the solution

[@cultist_pixel]

​@@khayalaliyev3519 i mean i saw the intended solution and it was kinda artificial, not something you come up on the spot. Coefficients are from positive R. BTW when i was solving it on that contest i tried a math induction and come to a form P(n+1)(2023)=Pn(2023)*F(b,c) where b and c were 2 indepentent numbers. I then took 2 different derivatives from F, set them equal to 0, crossed their solutions and proved that F(b0, c0) was greater than 2024. I didn't at the time check if that point was the minimum and the contest judges also didn't check. So it was counted as solved correctly even tho i later checked and (b0, c0) was a saddle point.

[@thomaslangbein297]

In the general case, it’s obvious and easy to prove, that it’s wrong. You can chose the a(i) big enough, depending on x, to make it work in the special case.

[@florentvaladier5099]

At 6:35 you can observe a pattern, sum((a+b)/(a²*b²)) then you can do partial fraction decomposition on it and see the telescopic sum

[@rishabsaini8347]

I'm stuck between programming and studying higher mathematics, this video helped me revisit some the topics I haven't touched in months.

[@elephantdinosaur2284]

If you replace perfect square numbers with another property like cubes, factorials, pell numbers, non-prime numbers you still get sum_n 1/(a_n * b_n) = 1. It really comes down to the fact that a_n & b_n are well-defined and non-zero for the property and includes 1. So something like limsup of the sequence of numbers with the property being infinite would be enough.

[@1tryhxrd.567]

you can use a double sum from the beginning without any ''drawing'' and you will get your result in no time

[@UltraMaXAtAXX]

I came here to escape my calculus II duties. Then partial fractions happened. Thanks, Mike.

[@aweebthatlovesmath4220]

What i like about the solution is that it can be generalised to perfect n powers and the answer will be 1-1/k^n+(k^n-l)/(k+1)^n where k is largest nth power less then l and l is the number of terms ( in this case n=2 , l=600).

[@udic01]

11:00-15:20 is a waste of time in this case. In 5:17 you explain why it is 2m+1. Which means that (2n+1)/[n^2(n+1)^2]=[(n+1)^2-n^2]/[n^2(n+1)^2]

[@bndrcr82a08e349g]

Nice exercise

[@zacknattack]

the thumbnail uses the wrong signs, using < instead of = instead of >

[@Animehater]

Did you use repertoire method to find A,B,C and D?

[@minwithoutintroduction]

نتيجة جميلة

[@lisandro73]

So when n tends to infinity the sum is just 1?

[@DrR0BERT]

Why oh why is partial fractions done this way? It pains me to see this tedious and laborious approach to the decomposition. It worked out here ok, but why not use the coverup/Heavyside method. This would instantly get B and D. Clearing fractions and moving the known terms (the terms without either A or C) to the other side and simplify. Saves on so much time and work.

[@gesucristo0]

What would be the limit as n tends to infinity?

[@davidecoli3108]

You can use the telescopic decomposition that Michael found at 15:54 and generalize it to a generic N. Basically you would find that the sum up to N is 1 - 1/(N+1)^2. Therefore the limit as N tends to infinity is just 1

[@landsgevaer]

@@davidecoli3108 You mean it can be generalized to a generic n of the form n = k*(k+1) (here, for 600, k = 24), and then the sum equals 1-1/n. Limit remains 1 indeed.

[@szymonraczkowski9690]

cool

[@joelneely]

Should be “largest perfect square at most n”

[@gp-ht7ug]

Cool

[@hugh081]

The key for me to solve this was to realise that (2n+1)/((n^2)(n+1)^2) was (1/n^2)-(1/(n+1)^2). I can't believe it took me a whole minute to remember that. Not really sure why he felt the need to decompose it formally since it's quite obvious, and after that the problem is trivial, but nice for n values which are equal to (k^2+(k+1)^2-1)/2, such as 600, but also 2, 6, 12, 20 etc. work nicely

[@titan1235813]

FFFFFFFFIIIIIIIIRRRRRRSSSSSTTTT! And first like and third view 😬👍🏻