Good day,
Here is an approximate formula:
s(n) = 38*n^3 - 222*n^2 + 404*n + 2^(2^n) - 218
V.Astanoff
Just by elementary-school arithmetic and by securing exponentially increasing quantities
of pencil, paper and eraser as you proceed towards infinity.
Happy studies: Johan E. Mebius
The general term is apparently 2^(2^n), n=0,1,2,... so, since each
term is exponentially greater than the previous one, the sequence of
partial sums is asymptotic to the final term in the partial sum.
(asymptotic in the sense that the ratio of the last term to the
partial sum becomes arbitrarily close to 1). Obviously the full series
diverges and has no sum.
Details are left as an exercise to the OP.
YOUR problem is in typing and proofreading, and you can solve your
problem by reading what you type, and by typing more slowly BEFORE you
press SEND.
Your last term should be 65536 = 2^16, in order for the terms of your
sequence to be 2^(2^i) for i=0 to infinity.
>> I want to find the sum of 2, 4, 16, 256, 65535, ..
>> Could you tell me how I can solve this problem?
>YOUR problem is in typing and proofreading, and you can solve your
>problem by reading what you type, and by typing more slowly BEFORE you
>press SEND.
I think you're the one with the problem here.
-- Richard
--
:wq
I think you're the one with the problem here. Right back at you.
I don't think so. First question is whether the OP means 65536 instead
of 65535. And if so, see:
http://www.research.att.com/~njas/sequences/A001146
--Lynn
no its you neilist.
> On Feb 22, 7:07 am, kasum...@gmail.com wrote:
> > I want to find the sum of 2, 4, 16, 256, 65535, ..
> > Could you tell me how I can solve this problem?
>
> The general term is apparently 2^(2^n), n=0,1,2,...
> so, since each
> term is exponentially greater than the previous one,
> the sequence of
> partial sums is asymptotic to the final term in the
> partial sum.
> (asymptotic in the sense that the ratio of the last
> term to the
> partial sum becomes arbitrarily close to 1).
trivial.
> Obviously the full series
> diverges and has no sum.
perhaps the OP wonders about a summability method.
>
> Details are left as an exercise to the OP.
what details ?? what exercise ??
regards
tommy1729
Tommy1729 still hasn't resolved his mental problems regarding
integrals.
On August 28, 2007, 2:35 PM, Tommy1729 wrote:
"integers ...-2,-1,0,1,2,...
integrals 1,2,3,4,...
natural numbers 0,1,2,3,... "
Hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha!
Hey Tommy, here's an "integral" for you: 69
Hahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahahaha!