# Can anyone remind me how to rewrite a summation?

1. Apr 7, 2017

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"Given: $sin(t)=Σ\frac{(-1)^nt^{2n+1}}{(2n+1)!}$
Prove: $L[sin(t)]=\frac{1}{s^2+1}$."

2. Relevant equations
$∑ar^k=\frac{a}{1-r}$

3. The attempt at a solution
$L[sin(t)]=L[Σ\frac{(-1)^nt^{2n+1}}{(2n+1)!}]=L[t]-\frac{1}{3!}L[t^3]+\frac{1}{5!}L[t^5]+...+\frac{1}{k!}L[t^k]-\frac{1}{(k+2)!}L[t^{k+2}]$
$L[sin(t)]=\frac{1}{s^2}-\frac{1}{s^4}+\frac{1}{s^6}+...+\frac{1}{s^{k+1}}-\frac{1}{s^{k+3}}=\sum_{i=1}^\infty (-1)^{i+1}(\frac{1}{s^{2i}})$
$\sum_{i=1}^\infty (-1)^{i+1}(\frac{1}{s^{2i}})=\frac{-1}{1-s^2}=\frac{1}{s^2-1}$

I'm stuck on primarily the last part. How would I get from the summation above to the Laplace transform of $sin(t)$?

Last edited: Apr 7, 2017
2. Apr 7, 2017

### Eclair_de_XII

Hold on a second... $\sum_{i=1}^\infty (-1)^{i+1}(\frac{1}{s^{2i}})=-\sum_{i=1}^\infty 1(-\frac{1}{s^2})^i=-\frac{1}{s^{-2}+1}$

Something's still not right, here...

Last edited: Apr 7, 2017
3. Apr 11, 2017

### Staff: Mentor

You forgot to set the limits here.

4. Apr 13, 2017

### Eclair_de_XII

$\sum_{k=0}^\infty ar^k=\frac{a}{1-r}$

In any case, I had to turn in the homework today, so I don't really need to solve this problem anymore.

5. Apr 13, 2017

### Staff: Mentor

For others stumbling on the thread: this sum starts from 0, while the one from the Laplace transform starts at 1. Taking this into account solves the problem.

@Eclair_de_XII: It is actually best not to make the first reply to your thread as it removes it from the unanswered thread list, and many helpers will miss it.

6. Apr 13, 2017

### Eclair_de_XII

I see. I will keep that in mind for the future. Thank you.