What is the summation of (1/n^3) as n is from 1 to infinity?? How do you arrive at the answer?

Answer 1

This sum equals Riemann's zeta(3), and it is also called Apery's constant because he showed it to be irrational (in 1979). No corresponding result is known for any odd number other than 3 (though it has been known, since 2001, that zeta(k) is irrational for infinitely many odd k, and that at least one of zeta(5), zeta(7), zeta(9) and zeta(11) is irrational).
For *even* k, zeta(k) is a rational multiple of pi^k; the coefficient is the abs. value of the k-th Bernoulli number times 2^(k-1)/k! (Euler, 1739).

Answer 2

There are a lot of series in my mathematical handbook, but unfortunately, this is not one of them. Most of them involve pi. For example, the summation of 1 / n^2 = pi^2 / 6.
If you can't find it on an internet search, then it probably either can't be done, or can be done, but no one has discovered it yet. All I can suggest is that you run a little program as I did and find out the answer to as many decimals as your computer will allow. It does seem to be converging and I'm sure there are many people out there who can prove or disprove that.

I summed to n = 208030, after which, I could get no more accuracy than 16 digits.
The summation came to 1.202056903150316.

Answer 3

you're complicated 2 distinctive tests. The attempt you're pertaining to says that if the cut back is nonzero, then the sequence ought to diverge. If the cut back is 0, then the attempt is inconclusive and bigger sorting out must be executed. for that reason the cut back as n is going to infinity a million/(n(n+a million)) is 0 so the attempt is inconclusive. so which you ought to save on with yet another attempt. the right attempt for that reason may be the assessment attempt and you ought to learn it with a million/n^2. considering that a million/n^2 is below a million/(n^2+n) (because of the fact the denominator will become larger so the whole fraction is smaller), and we additionally know that a million/n^2 converges (via the p-sequence attempt because of the fact p=2, in certainty it converges to pi^2/6), all of us know that a million/(n^2+n) additionally ought to converge. in certainty, a million/(n^2+n) converges to a million considering which you could decompose the fraction and get an alternating sequence. in short, this sequence would not pass against the attempt for divergence.

Answer 4

When I add the first 100 terms I get 1.202007401, and it looks to be converging.

Answer 5

you cant arrive at one answer, the answer like the pronumeral n is infinite