d***@gmail.com

2017-06-16 02:07:56 UTC

what do infinite sum x^k/(3k+2) as k ->infinity equal to?

reference math handbook

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reference math handbook

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d***@gmail.com

2017-06-16 02:07:56 UTC

what do infinite sum x^k/(3k+2) as k ->infinity equal to?

reference math handbook

www.mathHandbook.com

reference math handbook

www.mathHandbook.com

William Elliot

2017-06-16 02:37:52 UTC

Huh? Are you asking if sum(k=1,oo) x^k/(3k + 2) converges?

d***@gmail.com

2017-06-16 05:05:49 UTC

yes. what is euqal to =?

reference math handbook

www.mathHandbook.com

reference math handbook

www.mathHandbook.com

m***@wp.pl

2017-06-16 08:58:42 UTC

yes. what is euqal to =?

such things you should avoid mathematical problems.

m***@wp.pl

2017-06-16 09:03:27 UTC

yes. what is euqal to =?

such things you should avoid mathematical problems.

David Petry

2017-06-16 08:22:22 UTC

what do infinite sum x^k/(3k+2) as k ->infinity equal to?

reference math handbook

www.mathHandbook.com

Does your "mathHandbook" answer the question?

d***@gmail.com

2017-06-17 00:15:52 UTC

what do infinite sum x^k/(3k+2) as k ->infinity equal to?

reference math handbook

www.mathHandbook.com

what is your result?

input your formula

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

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Roland Franzius

2017-06-17 05:59:21 UTC

what do infinite sum x^k/(3k+2) as k ->infinity equal to?

reference math handbook

www.mathHandbook.com

what is your result?

input your formula

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

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

Roland Franzius

Markus Klyver

2017-06-16 09:01:12 UTC

It will only converge if abs(x) =< 1.

Jim Burns

2017-06-16 12:39:56 UTC

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 + ...

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 + ...

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