# Sum of a series that tends to infinity

• Physics lover
In summary: Yes. And what is (1-1/2)-1/2?In summary, 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 S=1/4+(1/1!)(1/4)^2+(1.3/2!)(1/4)^3-.
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?

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

How does one read ## 1.3.5/3!##?

WWGD and SammyS
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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chaksome and nuuskur
timetraveller123 said:
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.

nuuskur said:
How does one read ## 1.3.5/3!##?
It is (1.3.5)##/##3!

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Physics lover said:
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

timetraveller123 said:
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!

Physics lover said:
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:
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.

mjc123 said:
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.

Physics lover said:
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?

mjc123 said:
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

mjc123 said:
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?

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

SammyS

## 1. What is a series that tends to infinity?

A series that tends to infinity is a mathematical concept where the sum of the terms in a series becomes infinitely large as the number of terms increases. This means that the series does not have a finite sum and will continue to increase without ever reaching a limit.

## 2. How do you calculate the sum of a series that tends to infinity?

The sum of a series that tends to infinity cannot be calculated exactly since it does not have a finite sum. However, there are certain techniques and formulas such as the limit comparison test and the integral test that can be used to estimate the sum of such series.

## 3. Can a series that tends to infinity be convergent?

No, a series that tends to infinity cannot be convergent. Convergent series have a finite sum, whereas a series that tends to infinity has an infinite sum. A series can either be convergent or divergent, and a series that tends to infinity would fall under the category of divergent series.

## 4. What is the importance of understanding series that tend to infinity?

Understanding series that tend to infinity is crucial in many areas of mathematics, physics, and engineering. These series often arise in real-world applications and can help in predicting the behavior of systems that grow without bound, such as population growth, interest rates, and radioactive decay.

## 5. Are there any real-life examples of series that tend to infinity?

Yes, there are many real-life examples of series that tend to infinity. One common example is the compound interest formula, where the amount of money in an account grows exponentially over time. Other examples include the growth of bacteria in a petri dish, the temperature of a cooling object, and the number of bacteria in a colony.

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