# Summation Verification

1. Oct 24, 2014

### MathewsMD

Hi,

I'm just trying to evaluate a series and would just appreciate if someone could either verify or correct me work.

Essentially, I have a series that I've produced:

-[(t^2)/2 + (t^5)/(2x5) + (t^8)/(2x5x8) + ...]

= - *sum from n = 0 to infinity* [(t^(3n+2))/(3n+2)!] = -e^t

Sorry for the poor syntax. I'm just in a slight rush here and can hopefully fix it up sooner rather than later. Regardless, i essentially get -e^t as my answer, and if someone could verify if this is correct, that would be very helpful. Thank you!

2. Oct 24, 2014

### Staff: Mentor

???
How do you figure that your series equals -et?
The Maclaurin series for -et is -(1 + t + t2/2! + ... + tn/n! + ...)

3. Oct 24, 2014

### MathewsMD

Yes, it seems very odd and certainly not equivalent. I tried simplifying the series, but I only got to -t2 sum [t^3n/(3n+2)! But I can't quite further simplify it from here to put it in a form that seems expressible in terms of a function of e^rt. Any hints? Is there any way for me to possibly simplify the factorial in the denominator? If I change the summation from n = 0 to n = 1 (and still to infinity), this would allow it to become t^(3n - 3)/(3n - 1)!, but that still doesn't really help....

4. Oct 24, 2014

### Staff: Mentor

What's the problem you're trying to solve?

5. Oct 24, 2014

### MathewsMD

There's actually no specific problem...

I essentially have this series (that was found in another problem) and although there are other representations which could help me solve this, i was wondering if there was a way for me to transform this into a function of e^rt....I don't quite see any ways to do so, but would welcome any suggestions or methods. I don't quite see how having t^(3n) terms in the sum can allow for the sum to be expressed as a function of e^rt.
I was also wondering if there are other ways to manipulate the denominator factorial to help make the sum become another expression.

6. Oct 25, 2014

### lurflurf

7. Oct 25, 2014

### lurflurf

It is easy if you know about complex numbers
let
$$A=\sum_{k=0}^\infty \frac{x^{3k}}{(3k)!} \\ B=\sum_{k=0}^\infty \frac{x^{3k+1}}{(3k+1)!} \\ C=\sum_{k=0}^\infty \frac{x^{3k+2}}{(3k+2)!} \\ t=-\tfrac{1}{2}(1+i\sqrt{3}) \text{then solve for C using}\\ e^x=A+B+C e^{t x}=A+t B+t^2 C \\e^{t^2 x}=A+t^2 B+t C$$