# Explain this equality?

1. Apr 7, 2014

### WK95

1. The problem statement, all variables and given/known data
$\sum\limits_{n=1}^ \infty \frac{5^{n}}{n!} = \sum\limits_{n=0}^ \infty \frac{5^{n}}{n!} -1$

2. Relevant equations

3. The attempt at a solution

How is this equality true? How does one get from the first to the second equation?

2. Apr 7, 2014

### micromass

Staff Emeritus
$$(a_1 + a_2 + a_3 .... ) = (a_0 + a_1 + a_2 + a_3 + ...) - a_0$$

3. Apr 7, 2014

### phyzguy

What is the first term in $$\sum_0^\infty \frac{5^n}{n!}$$ ?

4. Apr 7, 2014

### Jilang

Are you sure that the sum on the RHS is not between 2 and infinity?

5. Apr 7, 2014

I'm certain.

6. Apr 7, 2014

### WK95

Thanks for the help. I get it now.

7. Apr 7, 2014

### Jilang

Yes I get it too now! It's a 5 on the RHS not a 1, LOL!

8. Apr 7, 2014

### phyzguy

What do you mean? 5^0 = 1, 0! = 1.

9. Apr 8, 2014

### Jilang

You need both to be true.

10. Apr 8, 2014

### micromass

Staff Emeritus
What? The equality in the OP is entirely correct. It doesn't need to be a 5 or anything.

11. Apr 8, 2014

### Staff: Mentor

What's the first term in the series on the right side? Hint: It's NOT 5.
???

12. Apr 8, 2014

### Staff: Mentor

You're missing the point of this problem, which is strictly about manipulating the indexes of a summation. It's not about whether the series converges or not.

13. Apr 8, 2014

### Jilang

Yes and you need that 5^0=1 and 0!=1.

14. Apr 8, 2014

### Staff: Mentor

These are true by definition. It wasn't clear from your previous comment about 5 on the right side, that you understood that both 5^0 and 0! were equal to 1.

15. Apr 8, 2014

### Jilang

Thanks for checking, I'm fine with it now though the 0! had me stumped for a while!