There is a slight flaw in argumentation. The fact that the series is bounded from below does not automatically mean that it's convergent. The sine function is bounded too, but it does not converge. The reason why it's convergent is that it is bounded from above and increasing (per your argument that each individual element in the sum is greater than zero).

@vinko8237

Ай бұрын

It is bounded, and every term is positive, so the sum is strictly increasing, and limited from above. That is enough

@DrBarker

Ай бұрын

Very good point! This could have been explained much more clearly - every term is positive, so the important property is that the sum is increasing, rather than just that it is bounded from below.

@boguslawszostak1784

Ай бұрын

@@DrBarker we have in 1:40 Sum =1/(sqrt(x^4+1)+x^2)>0

I guess the relevance of positive terms isn't just that the sum can't be negative, but also that the partial sums keep increasing towards either infinity or an upper bound. After all, a general series could also diverge by oscillation.

@skylardeslypere9909

Ай бұрын

Exactly

@DrBarker

Ай бұрын

Very good point - this could have been explained much better!

Absolutely loving the 1970s Open University vibe. Proper rigour, not many frills and even non distracting delivery. Can I suggest sideburns, brown suit and a kipper tie and the internet is yours for the taking.

The quick and dirty way would go something like this: sqrt(n^4 +1) = n^2(sqrt(1 + 1/n^4) ~ n^2(1+1/(2n^4)) = n^2 + 1/(2n^2). The whole expression therefore ~ 1/(2n^2) …

My first idea was multiplying by (sqrt(n^4+1)+n^2)/(sqrt(n^4+1)+n^2) then comparison test Series 1/n^2 is convergent so given series also is convergent Based on that he wrote in the description it is good way

@JohnCavendish-ql4jc

Ай бұрын

My method too.

Very nice! Alternatively, set (n+e)^4=n^4+1, from which it follows that 0 Therefore sqrt(n^4+1) -n^2 Knowing that Σ1/n^k exists for k>1, it follows that the series is convergent. Using the known values for ζ(2) and ζ(6) it follows that the value is smaller than π^2/12 + π^6/15120 < 0.89

@DrBarker

Ай бұрын

This is a very nice alternative!

@robertveith6383

Ай бұрын

Your second and third lines have errors because they are missing required grouping symbols. For example, 1/4n^3 *means* (1/4)n^3 by the Order of Operations. So, you need to have written 0 < e < 1/(4n^3) to express what you intended. And so on.

Lovely. A nice quick derivation with one eye on the election, I guess.😆

Let (n^2 + eta)^2 = n^4 + 1 It follows that eta^2 + (2n^2)eta = 1 It follows that eta must be smaller than 1/(2 * n^2) if we have (eta^2 + (2n^2)eta) = 1 It can also be noted in passing that eta^2 will be smaller than 1/(4* n^4) Accordingly sqrt(n^4 + 1) = n^2 + era So, sqrt(n^4 + 1) - n^2 = eta And since eta < 1/n^2, the sum of the series must converge.

Just use taylor series expansion on (n^4+x)^1/2 for x=1。n^2 cancel out and the rest of the terms are smaller than a geometric sum of 1/(n^2)^k, k>0.

Of course it's convergent. It's (n²+1/(2n²)+O(1/n⁶)-n²), so its O(n^-2).

3 ton Elephant in the room is the sum of the squares of the reciprocals famously converges to PI^2/6. Your way is instructive in the sense is proves convergence without finding the value series converges to.

For The Completion!

The sum is approximately √2 - 5 +√17- (5/8) + (π^2/12) 😂

Just multiply by root(n^4+1) +n^2 and divide by the same. You get 1 over root(n^4 +1) + n^2 that clearly goes to zero.

@pedroteran5885

Ай бұрын

That's a necessary but not sufficient condition for the series to converge.

I haven't watched the video but i think comparison with 1/n² would work well

But what is the sum?

@LuizPoublan

Ай бұрын

That's another beast entirely

@dalibormaksimovic6399

Ай бұрын

I calculated it, its around pi^2 /12

@user-cd9dd1mx4n

Ай бұрын

​@dalibormaksimovic6399 No. It is around 0.734572122454611 But pi^2/12 is around 0.822467 Percentage error is about 11.97%. Hence not a good approximation.

@dalibormaksimovic6399

Ай бұрын

@@user-cd9dd1mx4n I know, when I just ignored in expansion everything after n^4

@user-cd9dd1mx4n

Ай бұрын

@@dalibormaksimovic6399 Why would you ignore, if that will lead to a significant error (above 10%)? I may ignore insignificant terms, only when my final result has an error 2% at most (depending upon my application). Generally an error of 10% is huge.

I still do not get the point of determining convergence without an actual sum. It seems to me like and empty game. Sex without orgasm, food without swalowing, vodka without alcohol... What is the sum and how to find it? Otherwise it is meaningles in my opinion.

what about the value?

@seitanarchist

Ай бұрын

That is likely very difficult, if not impossible, to figure out. I would bet that it is unknown. Even computing the much more basic sum of the terms 1/n^2 requires somewhat sophisticated machinery.

I'm not watching the video. At a glance, multiplying top and bottom by the conjugate yields 1/n^2+more which converges by the p-test. Somebody hit me up if that's not the end result, or how it was done. Pls and thnx.

sqrt(3)? Edit: no it's slightly larger than this when N is 200.. Edit again: If you take n from 0 to +inf, the sum is sqrt(3) + some change, approx. sqrt(3.01)

Got lost in the first step😂😂

I checked out by mental calculation and it seems to converge towards zero and it is not too difficult to see why and one could show per complete induction that the larger the number to get the root of the smaller gets your result in that you got always a quadratic number plus one, which shreds you the decimal fractions in ever smaller pieces rooting them, so you end up converging towards zero. Le p'tit Daniel, Mama Christine I want to be with you making maths and burgers🐕🐕🐕🐕🐕

@robertveith6383

Ай бұрын

No, that is not logical.You are already starting out with a positive value in the summation when n = 1, and every term that is added is necessarily positive, so that the sum must be greater than zero.