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Eclair_de_XII

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## Homework Statement

"Given: ##sin(t)=Σ\frac{(-1)^nt^{2n+1}}{(2n+1)!}##

Prove: ##L[sin(t)]=\frac{1}{s^2+1}##."

## Homework Equations

##∑ar^k=\frac{a}{1-r}##

## 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)##?

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