hi can I ask if this series right or wrong. I found it with my calculations thanks

Dr.Peterson

Elite Member
when I try to check this summation, the result have ln(2) !
can anyone here know why is that happen ?!
I use wolframalpha.com
View attachment 29250
Why didn't you make the signs alternate, as in post 12?

And did you try the summation from the OP in WA?

New member
which post 12 you mean ?

Dr.Peterson

Elite Member
which post 12 you mean ?

Yours:
I can't define it.
I wrote how it supposed to be if we write it as an infinite summation.
we can't write this series as infinite series.
it's like this one in this photo , it's the minus side of this series and it's define for negative integers.
View attachment 29234
The signs there alternate.

New member
I see
and yes the sign is ulternate for negative integers side.
my post here is about if we but minus integer in the positive side of the equation.
for example if you put x=-5 in this positive side equation

you will get the answer having ln(2) according to wolfram language !!
why is that happen I don't know .

Dr.Peterson

Elite Member
I see
and yes the sign is ulternate for negative integers side.
my post here is about if we but minus integer in the positive side of the equation.
for example if you put x=-5 in this positive side equation
View attachment 29254

you will get the answer having ln(2) according to wolfram language !!
why is that happen I don't know .
But that's not what you put into WA! This has x+1, ... in the denominators, not x-1, ... as you showed.

You may include a link to your WA results, rather than an image, which may help us interact better.

New member
this equation have 2 formulas.
one for the positive integers which have (+) always.
and one for the negative integers which have alternative signs.

New member
if we take the positive side with the positive integers it will work and the same for the negative side.
I was trying to take the positive side for the negative integers such as (x=-5) but it gave me logarithm 2 always. !!

New member
hi
I found a simple proof for this infinite summation and you can check it in this picture.

New member
this one is without mistakes...

New member
and this one to make it a continued fraction..

New member
so you can write it ..
1/x=(1/x1)(0!+(1/x2)(1!+(1/x3)(2!+(1/x4)(3!+(1/x5)(4!+(1/x6)(5!+...∞

where
x1=x+1
x2=x+2
x3=x+3 ...and so on.