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

Relevant 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?

 

[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.

 

[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.

 

[Physics lover]

How does one read $1.3.5/3!$?

It is (1.3.5)$/$3!

 

[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]

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 $1 \times 3$ .

 

[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.

 

[mjc123]

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?