Convergence of an Interesting Series 

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).

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.

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.

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

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) …

Very nice! Alternatively, set (n+e)^4=n^4+1, from which it follows that 0 < e < 1/4n^3. Therefore sqrt(n^4+1) -n^2 < (n+ 1/4n^3)^2 - n^2 = 1/2n^2+1/16n^6 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

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.

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.

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

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.

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

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

For The Completion!

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.

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.

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)

But what is the sum?

what about the value?

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🐕🐕🐕🐕🐕