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

• stanli121
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).
stanli121

## Homework Statement

What is $$\sum_{n=0}^{\infty}{\frac{1}{(2n)!}}$$?

## Homework Equations

$$e^x$$ = $$\sum_{n=0}^{\infty}{\frac{x^n}{n!}}$$

## 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?

stanli121 said:

## Homework Statement

What is $$\sum_{n=0}^{\infty}{\frac{1}{(2n)!}}$$?

## Homework Equations

$$e^x$$ = $$\sum_{n=0}^{\infty}{\frac{x^n}{n!}}$$

## 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

have you treid working backwards from the answer to understand how they get there?

stanli121 said:

## Homework Statement

What is $$\sum_{n=0}^{\infty}{\frac{1}{(2n)!}}$$?

## Homework Equations

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

What x gives $(-1)^nx^n= (-x)^n= 1$?

## 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?

HallsofIvy said:
Perhaps more to the point
$$cos(x)= \sum_{n=0}^\infty \frac{(-1)^nx^n}{(2n)!}$$

What x gives $(-1)^nx^n= (-x)^n= 1$?

Um, actually
$$cos(x)= \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}$$
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

## 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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