Partial sum formula of a series by recursion

Eulero · 2017-01-09 05:57:13

Hi,
I haven't been here for a while, but now i'm back with something new. I found a formula that give the result of the partial sum of the series:
$FluxBB bbcode test$
For each k positive integer. With recursion i mean: do you want the partial sum formula for n=3? You need to know the partial sum formula for n=2 and for that you need partial sum formula for n=1;etc.
This is the formula:

$FluxBB bbcode test$

It works perfectly!
Before I publish the proof i really would like your judge:is it a useful formula? Or it's less interesting than i think?
I thank you for every answer.

bobbym · 2017-01-09 17:22:49

Hi;

That does indeed work for what I have tested but there is already a simple closed form for the first sum.

Does your solution have something special that makes it better for something? If so, what?

Eulero · 2017-01-09 23:06:48

Well, I never saw the question by this perspective. I think that my formula is simpler because doesn't need the knowledge of bernoulli numbers that are really hard to remember. And if k gets really high my formula is long because of the recursion but still practicable: i challenge everyone at remembering for a long time a big part of Bernoulli sequence. So should i post the proof or it's a naive work?

Last edited by Eulero (2017-01-09 23:07:38)

bobbym · 2017-01-10 00:47:01

Hi;

I would say go ahead and give it a shot. Where do you intend to publish?

Eulero · 2017-01-10 01:32:08

I think that i used the wrong word (I'm italian), with publish i mean post here. However this is my proof:
FluxBB bbcode test
Knowing that:
(excuse me the under subscript is 0)
We can rewrite the series so:

Now i notice that delta_n appears m-n times in the series:

We must work a little more for our formula:

QED

iamaditya · 2017-01-10 01:40:32

Hi, Eulero
The formula is too complicated to remember. It can be remembered much more easily by Implementing Faulhaber's formula. It says,
1ˣ +2ˣ +3ˣ +4ˣ ......nˣ =[1/(1+x)](aB₀nˣ⁺¹+bB₁nˣ+cB₂nˣ⁻¹............yBₓn), where Bₙ is the nth Bernoulli no. and a,b,c,.....y are the consecutive terms of (x+1)th row of Pascal's Triangle.
For e.g.
1¹¹+2¹¹+3¹¹+4¹¹.....7¹¹=(1/12)(1B₀n¹²+12B₁n¹¹+66B₂n¹⁰+220B₃n⁹+495B₄n⁸+792B₅n⁷+924B₆n⁶+792B₇n⁵+495B₈n⁴+220B₉n³+66B₁₀n²+12B₁₁n)
(1+B)ⁿ⁺¹-Bₙ₊₁=0
For e.g.
for,B₁ n=1 so
(1+B)²-B₂=0
⇒1+B₂+2B-B₂=0
⇒1=2B=0
⇒B=-0.5
However in this ( the formula for sums of powers )formula B₂=+0.5

Last edited by iamaditya (2017-01-10 01:41:52)

Math Is Fun Forum

#1 2017-01-09 05:57:13

Partial sum formula of a series by recursion

#2 2017-01-09 17:22:49

Re: Partial sum formula of a series by recursion

#3 2017-01-09 23:06:48

Re: Partial sum formula of a series by recursion

#4 2017-01-10 00:47:01

Re: Partial sum formula of a series by recursion

#5 2017-01-10 01:32:08

Re: Partial sum formula of a series by recursion

#6 2017-01-10 01:40:32

Re: Partial sum formula of a series by recursion

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