Discussion:
infinite sum
d***@gmail.com
2017-06-16 02:07:56 UTC
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what do infinite sum x^k/(3k+2) as k ->infinity equal to?

reference math handbook
www.mathHandbook.com
William Elliot
2017-06-16 02:37:52 UTC
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Post by d***@gmail.com
what do infinite sum x^k/(3k+2) as k ->infinity equal to?
Huh? Are you asking if sum(k=1,oo) x^k/(3k + 2) converges?
d***@gmail.com
2017-06-16 05:05:49 UTC
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Post by William Elliot
Post by d***@gmail.com
what do infinite sum x^k/(3k+2) as k ->infinity equal to?
Huh? Are you asking if sum(k=1,oo) x^k/(3k + 2) converges?
yes. what is euqal to =?

reference math handbook
www.mathHandbook.com
m***@wp.pl
2017-06-16 08:58:42 UTC
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Post by d***@gmail.com
Post by William Elliot
Post by d***@gmail.com
what do infinite sum x^k/(3k+2) as k ->infinity equal to?
Huh? Are you asking if sum(k=1,oo) x^k/(3k + 2) converges?
yes. what is euqal to =?
It's aleph0. Sorry to say, but if You have to ask
such things you should avoid mathematical problems.
m***@wp.pl
2017-06-16 09:03:27 UTC
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Post by m***@wp.pl
Post by d***@gmail.com
Post by William Elliot
Post by d***@gmail.com
what do infinite sum x^k/(3k+2) as k ->infinity equal to?
Huh? Are you asking if sum(k=1,oo) x^k/(3k + 2) converges?
yes. what is euqal to =?
It's aleph0. Sorry to say, but if You have to ask
such things you should avoid mathematical problems.
A nice compromitation for me. Sorry for that.
David Petry
2017-06-16 08:22:22 UTC
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Post by d***@gmail.com
what do infinite sum x^k/(3k+2) as k ->infinity equal to?
reference math handbook
www.mathHandbook.com
WolframAlpha gives the answer in terms of hypergeometric functions. Doing it by hand gives an answer in terms of logarithms.

d***@gmail.com
2017-06-17 00:15:52 UTC
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Post by David Petry
Post by d***@gmail.com
what do infinite sum x^k/(3k+2) as k ->infinity equal to?
reference math handbook
www.mathHandbook.com
WolframAlpha gives the answer in terms of hypergeometric functions.
it is too complicated. there is a simple result but wolfram cannot do it.
Post by David Petry
Doing it by hand gives an answer in terms of logarithms.
Post by David Petry
x^k/(3k+2)
into website www.mathHandbook.com, click the sum button for answer.
http://www.mathhandbook.com/input/?guess=infsum%28x%5Ek%2F%283k%2B2%29%29&inp=x%5Ek%2F%283k%2B2%29

reference math handbook
www.mathHandbook.com
Roland Franzius
2017-06-17 05:59:21 UTC
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Post by d***@gmail.com
Post by David Petry
Post by d***@gmail.com
what do infinite sum x^k/(3k+2) as k ->infinity equal to?
reference math handbook
www.mathHandbook.com
WolframAlpha gives the answer in terms of hypergeometric functions.
it is too complicated. there is a simple result but wolfram cannot do it.
Post by David Petry
Doing it by hand gives an answer in terms of logarithms.
Post by David Petry
x^k/(3k+2)
into website www.mathHandbook.com, click the sum button for answer.
http://www.mathhandbook.com/input/?guess=infsum%28x%5Ek%2F%283k%2B2%29%29&inp=x%5Ek%2F%283k%2B2%29
This gives the most superflous answer a computer can give in such a
case. Any Taylor series Phi(x,...) = sum_k a_k x^k with a simple
multiply fractional linear regression formula of the coefficients a_k->
a_(k+1) is called a Gauss hypergeometric function as a generalization of
the geometric series (1-x)^-1 = sum_k x^k.

So, what is gained to display a clear and simple infinite sum as a
member of the family of hypergeometric functions with a special
selection of its many parameters.

(answer: a one-liner in Mathematica eg to retrieve the coeffients in the
recursion formula for the coefients. Nobody with the least mathematical
skill uses a computer algebra system for such a nonsense)

This is the "Schwachsinn"
(http://dict.leo.org/englisch-deutsch/schwachsinn) of the form we mostly
find in help files from other idiots in the computer and social sciences
branches, that explicate unknown terms in contexts of more
even-more-unknown terms.
--
Roland Franzius
Markus Klyver
2017-06-16 09:01:12 UTC
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It will only converge if abs(x) =< 1.
Jim Burns
2017-06-16 12:39:56 UTC
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Post by d***@gmail.com
what do infinite sum x^k/(3k+2) as k ->infinity equal to?
I'll call that sum S, assuming k >= 0.

Let x = y^3.
The sum S
1/2 + x/5 + x^2/8 + ...
equals
(1/y^2)( y^2/2 + y^5/5 + y^8/8 + ... )

Note that
y/1 + y^2/2 + y^3/3 + y^4/4 + y^5/5 + ...
equals
ln(1/(1 - y))

for |y| =< 1 & ~(y = 1)

Define a clock sequence c(k) such that
c(1), c(2), c(3), c(4), c(5), c(6), ...
equals
0, 1, 0, 0, 1, 0, ...

The sum
c(1)*y/1 + c(2)*y^2/2 + c(3)*y^3/3 + ...
equals
y^2/2 + y^5/5 + y^8/8 + ...

Let e be a solution of u^3 = 1 other than u = 1.
Thus
(e^3 - 1)(e - 1) = e^2 + e + 1 = 0
There are two possible solution. Let
e = - 1/2 + sqrt(3)/2*i
Define
c(k-1) = ( e^2k + e^k + 1 )/3
Then
c(1), c(2), c(3), c(4), c(5), c(6), ...
equals
0, 1, 0, 0, 1, 0, ...

If the following converges absolutely, we can
re-arrange the terms of
c(1)*y/1 + c(2)*y^2/2 + c(3)*y^3/3 + ...
as the sum of three infinite sums
(e^2/3)( (e^2*y)/1 + (e^2*y)^2/2 + ... )
+
(e/3)( (e*y)/1 + (e*y)^2/2 + (e*y)^3/3 + ... )
+
(1/3)( y/1 + y^2/2 + y^3/3 + y^4/4 + ... )

which equals
(e^2/3)*ln(1/(1 - e^2*y))
+ (e/3)*ln(1/(1 - e*y))
+ (1/3)*ln(1/(1 - y))

for |y| < 1 and e = - 1/2 + sqrt(3)/2*i

I will write cbrt(x) = x^(1/3)
and e = - 1/2 + sqrt(3)/2*i

Retracing our steps, we see that
1/(3*cbrt(x)^2)*
( e^2*ln(1 - e^2*cbrt(x))
+ e*ln(1 - e*cbrt(x))
+ ln(1 - cbrt(*x)) )

is the value of S, the sum
1/2 + x/5 + x^2/8 + ...