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Sum of a series that tends to infinity

Homework Statement
The sum of series ##1+(1/1!)(1/4)+(1.3/2!)(1/4)^2+(1.3.5/3!)(1/4)^3.....## to infinity is
Homework Equations
sum of infinite gp=##a/(1-r)##
Sum of infinite agp=##ab/(1-r)+dbr/(1-r)^2##
I tried by
##S=1+(1/1!)(1/4)+(1.3/2!)(1/4)^2+...##
##S/4=1/4+(1/1!)(1/4)^2+(1.3/2!)(1/4)^3..##
And then subtracting the two equations but i arrived at nothing What shall i do further?
 
i think this method might work the sum in question looks a lot like a mclaurin series
so maybe can you think of a function whose mclaurin series takes this form
since the product on top goes like 1,3,5,7 it suggests that the exponent on the function can't be integral so it has to some kind of fractional exponent . so which function can you think of for this form
edit: the mclaurin also needs to converge
 
511
336
How does one read ## 1.3.5/3!##?
 
57
30
Hi.
I just state the question and do some estimation
[tex]S=\sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!}(2r)^n <\sum_{n=0}^\infty (2r)^n =\frac{1}{1-2r}[/tex]
where r=1/4 and 0!!=(-1)!!=1

[tex]S=\frac{1}{\sqrt{1-2r}}[/tex]
We can confirm it by expanding RHS by 2r.
 
Last edited:
i think this method might work the sum in question looks a lot like a mclaurin series
so maybe can you think of a function whose mclaurin series takes this form
since the product on top goes like 1,3,5,7 it suggests that the exponent on the function can't be integral so it has to some kind of fractional exponent . so which function can you think of for this form
edit: the mclaurin also needs to converge
I am in high school and i can not use Mclaurin series.
 
It is (1.2.3)///3!
what do you mean so is each term of the form ##\frac{1.2.3.4...n}{n!}## or ##\frac{1.3.5...(2n-1)}{n!}##

edit never mind you were probably just giving an example i take it as the second one
 
what do you mean so is each term of the form ##\frac{1.2.3.4...n}{n!}## or ##\frac{1.3.5...(2n-1)}{n!}##

edit never mind you were probably just giving an example i take it as the second one
Soory that was a typing mistake.
It's(1.3.5)/3!
 

SammyS

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Homework Helper
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Homework Statement: The sum of series ##1+(1/1!)(1/4)+(1.3/2!)(1/4)^2+(1.3.5/3!)(1/4)^3.....## to infinity is
Homework Equations: sum of infinite gp=##a/(1-r)##
Sum of infinite agp=##ab/(1-r)+dbr/(1-r)^2##

I tried by
##S=1+(1/1!)(1/4)+(1.3/2!)(1/4)^2+...##
##S/4=1/4+(1/1!)(1/4)^2+(1.3/2!)(1/4)^3..##
And then subtracting the two equations but i arrived at nothing What shall i do further?
When you write ## 1.3 ##, I suppose that you mean ## 1## times ##3##, rather than ##13/10## .

In LaTeX, use \cdot to get ## 1 \cdot 3## , or use \times to get ## 1 \times 3## .
 
Last edited:

mjc123

Science Advisor
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Consider the expansion
(1+x)n = 1 + nx + n(n-1)x2/2! ...
Try thinking of a value of n such that n(n-1)(n-2)... gives 1.3.5... in the numerator.
 
Consider the expansion
(1+x)n = 1 + nx + n(n-1)x2/2! ...
Try thinking of a value of n such that n(n-1)(n-2)... gives 1.3.5... in the numerator.
if i put n=5/2,then i would get (1## . ##3## . ##5)/8
 
Please somebody help me to proceed further.
 

mjc123

Science Advisor
814
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if i put n=5/2,then i would get (1.. . 3.. . 5)/8
But only for that term. (Did you spot the ... ?) What might you start with that could give 1 then 1.3 then 1.3.5 then 1.3.5.7... in the numerator?
 
But only for that term. (Did you spot the ... ?) What might you start with that could give 1 then 1.3 then 1.3.5 then 1.3.5.7... in the numerator?
i think it's n=-3/2.
 

mjc123

Science Advisor
814
369
Try n = -1/2
 

mjc123

Science Advisor
814
369
Yes. And what is (1-1/2)-1/2?
 

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