simple proof of an infinite summation

You already started a thread for this, so it might make sense if you post your correction in the other thread(click) and abandon this one (perhaps a moderator will delete this thread)

FYI: It looks correct to me. Do you have a use in mind for it?
 
I have two comments.

First, I would call this a derivation, rather than a proof. A proper proof might use induction, using the same ideas you show here.

Second, you say nothing about convergence. Without that, this can't really be called true or false. When does it converge? Where is the proof?
 
Cubist said:
You already started a thread for this, so it might make sense if you post your correction in the other thread(click) and abandon this one (perhaps a moderator will delete this thread)

FYI: It looks correct to me. Do you have a use in mind for it?
Click to expand...
it's an infinite series but you can stop at any point you want to have a set of fractions from one given fraction.
for example
1/4=1/5+ 1/20. for 2 fractions. or
1/4=1/5+ 1/30+ 1/60. for 3 fractions or
1/4= 1/5 +1/30+1/105+1/140. for 4 fractions.
and you can get whatever numbers of fractions from 1/4.
 
Dr.Peterson said:
I have two comments.

First, I would call this a derivation, rather than a proof. A proper proof might use induction, using the same ideas you show here.

Second, you say nothing about convergence. Without that, this can't really be called true or false. When does it converge? Where is the proof?
Click to expand...

I didn't continue in adding fractions to reach the sufficient evidence of this continuation.
any way if you try it by yourself then you will get the point of it.
 
if you try wolfram.com you will get that this infinite series have a convergence always except for x=0
 
so how to make a proof that this series is converges for all Xs except 0 ?!
I didn't try this before
 
if we take the second parameter divided by first parameter:

Screenshot_٢٠٢١_١٠٢٢_١٥٥٤٣٠.png

Lim (1!/((x+1)(x+2)))/(0!/(x+1))
= Lim (1/(x+2)). which is less than 1 for all positive integers .
is this a fair proof that this series is converges ?!
 
or the summation for this series is ..

Screenshot_٢٠٢١_١٠١٤_١٥١٩٢٦.png

so we can take
Lim (f(x+1)/f(x))
so => Lim (n!(x+1)!/(n+x+1+1)!)/(n!x!/(n+x+1)!)
then ...
= (x+1)/(n+x+2)
so it's obvious that the result is less than 1 for positive integer also.
I think it's a fair proof for it's convergence.

and you can see that the limit for n→∞ is equal to zero.
Lim ((x+1)/(∞+x+2)) = zero.
 
Mohammad Hammad said:
it's an infinite series but you can stop at any point you want to have a set of fractions from one given fraction.
for example
1/4=1/5+ 1/20. for 2 fractions. or
1/4=1/5+ 1/30+ 1/60. for 3 fractions or
1/4= 1/5 +1/30+1/105+1/140. for 4 fractions.
and you can get whatever numbers of fractions from 1/4.
Click to expand...
I understand that you can turn a fraction into an exactly equivalent sum of fractions, but do you have a use in mind for this? Or, perhaps, this is just an interest of yours (which is absolutely fine)?

Have you considered that there may be more than one way to break a fraction down into the sum of two fractions. For example:-
1/24 = 1/48 + 1/48 = 1/42 + 1/56 = 1/40 + 1/60 = 1/36 + 1/72 = 1/33 + 1/88 = 1/32 + 1/96 = 1/30 + 1/120 = 1/28 + 1/168 = 1/27 + 1/216 = 1/26 + 1/312 = 1/25 + 1/600
 
it came out with me with many summations from a method I am testing.
so I am just asking if someone here have another point of view about it.
thanks for everyone who spend the time to see my threads here.
 
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