Solving Tricky Infinite Sum: What is \sum_{n=0}^{\infty}{\frac{1}{(2n)!}}?

In summary, the conversation was about finding the sum of the infinite series \sum_{n=0}^{\infty}{\frac{1}{(2n)!}} and how it relates to the series for e^x and cos(x). The answer was given as (1/2)(e+e^-1), but the person was struggling to understand where the e^-1 term came from. They were given hints to work backwards from the answer and to consider the series for cos(x).
  • #1
stanli121
12
0

Homework Statement



What is [tex]\sum_{n=0}^{\infty}{\frac{1}{(2n)!}}[/tex]?

Homework Equations


[tex]e^x[/tex] = [tex]\sum_{n=0}^{\infty}{\frac{x^n}{n!}}[/tex]


The Attempt at a Solution


I understand that the given series looks like the series for e^1 but I know that isn't the correct answer. The answer (without solution) was supplied as (1/2)(e+e^-1). I can't seem to figure out the extra e^-1 part. Help?
 
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  • #2
stanli121 said:

Homework Statement



What is [tex]\sum_{n=0}^{\infty}{\frac{1}{(2n)!}}[/tex]?

Homework Equations


[tex]e^x[/tex] = [tex]\sum_{n=0}^{\infty}{\frac{x^n}{n!}}[/tex]


The Attempt at a Solution


I understand that the given series looks like the series for e^1 but I know that isn't the correct answer. The answer (without solution) was supplied as (1/2)(e+e^-1). I can't seem to figure out the extra e^-1 part. Help?

Try writing out e (which equals e1) and e-1 as infinite sums, using the equation you have given.

Cheers -- sylas
 
  • #3
have you treid working backwards from the answer to understand how they get there?
 
  • #4
stanli121 said:

Homework Statement



What is [tex]\sum_{n=0}^{\infty}{\frac{1}{(2n)!}}[/tex]?

Homework Equations


[tex]e^x[/tex] = [tex]\sum_{n=0}^{\infty}{\frac{x^n}{n!}}[/tex]
Perhaps more to the point
[tex]cos(x)= \sum_{n=0}^\infty \frac{(-1)^nx^n}{(2n)!}[/tex]

What x gives [itex](-1)^nx^n= (-x)^n= 1[/itex]?


The Attempt at a Solution


I understand that the given series looks like the series for e^1 but I know that isn't the correct answer. The answer (without solution) was supplied as (1/2)(e+e^-1). I can't seem to figure out the extra e^-1 part. Help?
 
  • #5
HallsofIvy said:
Perhaps more to the point
[tex]cos(x)= \sum_{n=0}^\infty \frac{(-1)^nx^n}{(2n)!}[/tex]

What x gives [itex](-1)^nx^n= (-x)^n= 1[/itex]?

Um, actually
[tex]cos(x)= \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}[/tex]
This is not going to help. And let's not introduce the hyperbolic cos. He's got the answer, and now just needs to work back from there.

Cheers -- sylas
 

Related to Solving Tricky Infinite Sum: What is \sum_{n=0}^{\infty}{\frac{1}{(2n)!}}?

What is the value of the infinite sum?

The value of the infinite sum is approximately 1.6487212707.

What is the pattern of the terms in the sum?

The terms in the sum follow a factorial pattern, where the denominator increases by 2 with each term and the numerator remains constant at 1.

How is this infinite sum solved?

This infinite sum is solved using a mathematical technique called the Maclaurin series or Taylor series expansion.

Why is this infinite sum considered tricky?

This infinite sum is considered tricky because it involves an infinite number of terms and the terms do not approach zero as n approaches infinity, making it difficult to find an exact solution.

What is the significance of this infinite sum in mathematics?

This infinite sum is significant in mathematics because it is an example of a convergent series, meaning that the sum of the terms approaches a finite value as the number of terms increases. It is also used in calculus and other areas of mathematics to approximate and solve various problems.

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