I really dont know how to call those:

n

E i

i=1

 

in english, so forgive me for the title, anyway, I have come with a big question for you forum, I have lately been working with

 

n

E 1/c^i

i=1

 

where: c > 1, c is natural

 

working with those I think I may have found a general formula for any c (that satisfies the past requirements), I was wondering if you have ever heard of it and also if you have it, I would appreaciate.

Later I will post my thinking, procedure and conclusion, meanwhile, thank you for your help.

The operation is called summation, and the method to calculate the general formula has been known for some time, although I confess I'm not sure if I'd be up to discovering it on my own. Have a look here to check your answer.

 

Also, posting the same topic thrice won't help you at all; on the contrary, you risk getting banned for spam. So don't.

The general formula for a finite geometric series is [imath]\sum\limits_{i=0}^{n}r^i = \frac{r^{n+1}-1}{r-1}[/imath], it's pretty easy to subsitute in [imath]r=\frac{1}{c}[/imath]. I can't see why [imath]c \in \mathbb{N}[/imath] should be a requirement at all.

Blopa, thanks for stepping up and asking an important question about mathematics.

 

So you can recognize the summation notation as its inventor intended, take a look at the capital sigma, Σ. People may not recognize it as a capitol "S", but it is. It's an angular capital "S" with a "foot" (perhaps it was originally a fancy 'finishing stroke'). Remove the "foot" from the Σ, and you can see the angular "S".

 

This capital Greek "S" is used to represent "summation". The integral symbol (also called the "long s" by linguists), ∫, is an "S" that also stands for "summation", but in a different way (perhaps best addressed here in a future thread).

Surely the Greek letter Sigma (Σςσ) predates the Latin Ss; to describe a capital Sigma as an S with "originally a fancy finishing stroke" reverses the development of alphabets (semitic->phoenician->greek->latin).

Jesus, now I feel really stupid...

I see, and sorry, I didnt knew i posted it thrice (I've been having some connection problems =S, seriously I'm really ashamed for the spam and the trouble, sorry forum)

and, oh I see, thank you for the help, it seems I'm late for this discovery as well -.-

Anyway thanks for your help, and it seems I was correct =D

 

What I did:

 

I was thinking of a kind off "infinite circular ladder" in which each "step" became half of the last one (or also and "empty disc" which I started filling), and thought of the

 

S 1/2^i

 

to solve it, I later thought that it will aproximate infinitly to one, but always would be 1 - (1/2^n) which would become infinitely small, it would always be missing "one of the same size as the last step" so I wrote:

 

S 1/2^i = 1 - 1/2^i

 

the next day I wrote the formula and thought "hey, what if instead of a 2 I use a 3?"

 

I made 1/3 + 1/9 + 1/27 = (9+3+1)/27 = 13/27 and I was like "oh -.-", I tried it again adding one more "step", 1/3 + 1/9 + 1/27 + 1/81 = (27+9+3+1)/81 = 40/81

 

And then I noticed something, the nominators (upper part of fraction) were equivalent to (3^n - 1)/2; so I wrote:

 

S 1/3^i = (3^n - 1)/(2 · 3^n) = 1/2 - 1/(2 · 3^n)

 

I thought "this looks kinda similar, lets work with the 4"

 

S 1/4^i

 

1/4 + 1/16 + 1/64 = (16+4+1)/64 = 21/64 this time I was like "what is this?"

1/4 + 1/16 + 1/64 + 1/256 = (64+16+4+1)/256 = 85/256

 

So I tried to find a relationship between 21/64 and 85/256

I noticed they were similar, and then I divided (actually I dont know why I did it, luck?) 256/3 and I found out it was incredibly close to 85, so I made (256-1)/3 = 85, I was impressed and did the same with the other one (64-1)/3 = 21, I had found the relationship

 

S 1/4^i = (4^n-1)/(3 · 4^n) = 1/3 - 1/(3·4^n)

 

Then I noticed the 3 formulas were similar, and after analyzing how they "changed" I got to the conclusion:

 

S 1/c^i = 1/(c-1) - 1/(c-1)(c^n)

 

I tested it this morning with a program I wrote, it worked on more than 99% of the times with different values (16 decimals, mistakes were due to computer approximation; I put her to calculate it by "adding" and meanwhile by using my formula to see if they were the same)

 

Anyway thanks for your time forum =)

Edited December 2, 201015 yr by Blopa

?????????

the first step of solve is find the formula!

Jesus, now I feel really stupid...

 

You are not.

 

the first step of solve is find the formula!

 

He used the empirical method. I used to do so when I was young, and in several circumstances still do. It gives the great excitement of discovery.