# Sum of a series that tends to infinity

#### Physics lover

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?

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#### timetraveller123

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

#### nuuskur

How does one read $1.3.5/3!$?

#### mitochan

Hi.
I just state the question and do some estimation
$$S=\sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!}(2r)^n <\sum_{n=0}^\infty (2r)^n =\frac{1}{1-2r}$$
where r=1/4 and 0!!=(-1)!!=1

$$S=\frac{1}{\sqrt{1-2r}}$$
We can confirm it by expanding RHS by 2r.

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#### Physics lover

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.

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#### timetraveller123

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

#### Physics lover

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

Staff Emeritus
Homework Helper
Gold Member
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 $1 \times 3$ .

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#### mjc123

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.

#### Physics lover

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

#### Physics lover

Please somebody help me to proceed further.

#### mjc123

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?

#### Physics lover

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.

Try n = -1/2

#### Physics lover

Try n = -1/2
ok so i should put n=-1/2 and x=-1/2 and that gives me the sum.Am i correct?

#### mjc123

Yes. And what is (1-1/2)-1/2?

"Sum of a series that tends to infinity"

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