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Infinite series challenge.

Anonymous Sun 14 Nov 16:27:57 2021 No.13852091 View Reply Original Report

Quoted By: >>13852185 >>13852209 >>13852555 >>13852592 >>13854324 >>13856582

Find the sum

Anonymous

Anonymous Sun 14 Nov 2021 17:01:32 No.13852185 Report

Quoted By: >>13852245

>>13852091
Looks like something between 0.57 and 0.58.
Finding an exact formula will be very hard since it's a conditionally convergent series so you can't use any term rearrangement tricks.

Anonymous

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Anonymous Sun 14 Nov 2021 17:07:06 No.13852209 Report

Quoted By:

>>13852091
Nice homework problem. I'll try and sketch things out.
Consider the following functions
(1) (-1)^(m+1) / (m+n)n
(2) 1 / (m+n)n
(3) 1 / (1+n)n
(4) 1 / n^2.
You should be able to figure out what the sum (4) converges to, and then you can relate this back to (1), (2), and (3) by a series of inequalities (this should be very simple). Once you do that, you should be able to answer your question.

Anonymous

Anonymous Sun 14 Nov 2021 17:14:45 No.13852245 Report

Quoted By: >>13852374

>>13852185
Brainlet here: wtf
How does one even calculate this shit with infinity in it? And aside, what does this express? What's the use of this formula?

Anonymous

Anonymous Sun 14 Nov 2021 17:54:01 No.13852374 Report

Quoted By: >>13852385

>>13852245
Not him but that infinity means that there is an infinite amount of terms in the sum and sometimes you can sum an infinite amount of numbers and have a finite result. One way this could happen is by making every successive number smaller than the last one. Another way is summing positive numbers with negative numbers, which is known as alternate series. OP's series obeys both conditions but I don't know wtf it means tho

Anonymous

Anonymous Sun 14 Nov 2021 17:55:31 No.13852385 Report

Quoted By:

>>13852374
https://plus.maths.org/content/when-things-get-weird-infinite-sums
Found this interesting article and it fucks with my mind. Interesting concept, never heard or thought of it that precicely.

Anonymous

Anonymous Sun 14 Nov 2021 17:58:53 No.13852390 Report

Quoted By:

$\frac{1}{12}\left(\pi^2-6\log^2(2)\right)$

Anonymous

Anonymous Sun 14 Nov 2021 18:03:20 No.13852405 Report

Quoted By:

equivalent to the single sum 1/n^2/2^n
similar to how 1/n/2^n -> log(2)
and how 1/n^2 -> zeta(2)
this -> $\frac{\zeta(2) - \log(2)^{2}}{2}$

Anonymous

Anonymous Sun 14 Nov 2021 18:40:14 No.13852543 Report

Quoted By: >>13852564

Isn't this the correct answer?:
The inner sum diverges and this whole sum is undefined. It doesn't converge absolutely and by Riemann theorem the simple summands can be reordered into any given sum including +-inf.

Nah... correction.. the infinite sum is defined as a limit, and since the inner sum diverges, this sum's limit is the inner one's and is therefore divergent too.

Anonymous

Anonymous Sun 14 Nov 2021 18:43:15 No.13852555 Report

Quoted By:

>>13852091
how did this problem arise? what is the origin story here?

Anonymous

Anonymous Sun 14 Nov 2021 18:45:49 No.13852564 Report

Quoted By:

>>13852543
Fuck, I'm sorry. Thought -1 was to the power of n, not m.

Anonymous

Anonymous Sun 14 Nov 2021 18:55:23 No.13852592 Report

Quoted By: >>13852820 >>13856056

>>13852091
I can estimate it to within a trillionth decimal point through a basic script if that's acceptable :3

Anonymous

Anonymous Sun 14 Nov 2021 20:04:51 No.13852820 Report

Quoted By: >>13852825

>>13852592
what's the 999999999999th decimal digit

Anonymous

Anonymous Sun 14 Nov 2021 20:07:38 No.13852825 Report

Quoted By:

>>13852820
10% chance it's 3

Anonymous

Anonymous Mon 15 Nov 2021 05:12:59 No.13854303 Report

Quoted By: >>13854310

If the sum is infinite, how can that be found out?

Anonymous

Anonymous Mon 15 Nov 2021 05:17:38 No.13854310 Report

Quoted By:

>>13854303
it's like if you piss into a toilet it goes there

Mohan

Mohan Mon 15 Nov 2021 05:26:38 No.13854324 Report

Quoted By:

>>13852091
> homework chalange

Anonymous

Anonymous Mon 15 Nov 2021 17:31:11 No.13855915 Report

Quoted By: >>13856059 >>13856298 >>13856577

To solve this, do I need to integrate this?
$\int \frac{x^n}{1+x}dx$
How can that be done?

Anonymous

Anonymous Mon 15 Nov 2021 18:26:18 No.13856056 Report

Quoted By:

>>13852592
>within a trillionth decimal point
i hope you have terabytes of ram in your computer before you trying doing that

Anonymous

Anonymous Mon 15 Nov 2021 18:27:27 No.13856059 Report

Quoted By: >>13856256

>>13855915
>How can that be done?
integration by parts immediately comes to mind

Anonymous

Anonymous Mon 15 Nov 2021 19:04:45 No.13856196 Report

Quoted By:

Mr. Hak would probably know this. Dutchfag

Anonymous

Anonymous Mon 15 Nov 2021 19:25:01 No.13856256 Report

Quoted By:

>>13856059
do go on

Anonymous

Anonymous Mon 15 Nov 2021 19:42:17 No.13856298 Report

Quoted By: >>13856306 >>13856370

>>13855915
My 2nd try would be with partial fraction decomposition, my 1st and very possibly only one would be numerical integration

Anonymous

Anonymous Mon 15 Nov 2021 19:44:32 No.13856306 Report

Quoted By:

>>13856298
or is it not needed for calculating the sum? I obtained it from a power series for the inner sum.

Anonymous

Anonymous Mon 15 Nov 2021 20:02:30 No.13856370 Report

Quoted By:

>>13856298
>partial fraction decomposition
Ah yes, this somehow works, but is ugly as hell.

Anonymous

Anonymous Mon 15 Nov 2021 20:56:40 No.13856577 Report

Quoted By:

>>13855915
$\frac{x^n}{1-x} =\frac{1}{1-x} - \sum_{k=1}^{n}x^k$

Anonymous

Anonymous Mon 15 Nov 2021 20:59:17 No.13856582 Report

Quoted By:

>>13852091
i'm fairly sure this diverges, inner sum is alternating harmonic series plus f(n), where f(n) doesn't converge

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