# What are some sums of infinite series that are = to 'e'?

1. Dec 11, 2013

### mesa

We all know about the sum of the infinite series,

1 + 1/1! + 1/2! + 1/3! + ... to 1/inf! = e

What other series do we have that are equal to 'e'?

2. Dec 11, 2013

### Panphobia

3. Dec 11, 2013

### mesa

4. Dec 11, 2013

### Mandelbroth

I like $$e=\sum_{n\geq 0}f(n),$$ where $\displaystyle f(n)=\left\{\begin{matrix}e & \text{if } n=0 \\ 0 & \text{if } n\neq 0\end{matrix}\right.$. :tongue:

5. Dec 11, 2013

### mesa

That's different... :)

6. Dec 11, 2013

### HallsofIvy

Staff Emeritus
Personally, I prefer $$e=\sum_{n\geq 0}f(n),$$ where $\displaystyle f(n)=\left\{\begin{matrix}e & \text{if } n=1 \\ 0 & \text{if } n\neq 1\end{matrix}\right.$

7. Dec 11, 2013

### mesa

So is this all of them, as in if someone found something else outside of these then it's new to maths?

8. Dec 11, 2013

### Mandelbroth

Do you honestly think we know all of mathematics?

I'd venture to guess that there are at least SOME (*cough*SUM*cough*) other representations out there. :tongue:

9. Dec 11, 2013

### mesa

Nope, just trying to see if someone else has more information. Do you have recommendations on where to find out more on the subject such as books, websites, etc.?

10. Dec 11, 2013

### Jorriss

That list is not exhaustive. There is, as far as I know, not much interest in simply finding new infinite series that add up to 'e,' except perhaps if a new series converges much more rapidly than a previously known series.

11. Dec 11, 2013

### mesa

I would imagine that would be valuable from a computing standpoint. What's the current record, 10 trillion decimal places or something to that effect?

12. Dec 12, 2013

### Jorriss

In what sense valuable? I do not believe there is much interest in finding more decimal places to e beyond the novelty of having found new decimals - though someone more knowledgeable can certainly correct me.

13. Dec 12, 2013

### Stephen Tashi

Why not concentrate on finding infinite series that sum to $1$ and then multiply them by $e$? That would let you find series that sum to $\pi$ or $42$ or whatever.

14. Dec 12, 2013

### mesa

I find it remarkable we live in a time where we know a 1,000,000,000,000+ decimal places of 'e'. Euler and the like would probably be in awe of such an accomplishment.

Also I think figuring out new ways of calculating 'e' is important since it could help us gain new insight and lead to better maths in the future (much like it has proven to do so in the past).

15. Dec 14, 2013

### mesa

Brilliant!

$$e=e*2\sum_{n= 2}^{∞}(n-1)/((2)n!)$$
$$42=42*2\sum_{n= 2}^{∞}(n-1)/((2)n!)$$
$$pi=pi*2\sum_{n= 2}^{∞}(n-1)/((2)n!)$$
$$llama=llama*2\sum_{n= 2}^{∞}(n-1)/((2)n!)$$

Yes I 'cheated' with the 2 multiplier but so far I have only been able to derive an infinite sum to 1/2 and finals are calling....
These things fun!