How Can I Calculate the Cauchy Sum of a Taylor Polynomial in Maple?

[Simfish]

So...

I want to find the Cauchy sum of the Taylor polynomial of \exp x \sin x. I know how to do this with maple, which only requires the command
taylor(sin(x)*exp(x), x = 0, n). I can also try the good old f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+\cdots formula, but that isn't the learning objective. But I want to see if I did the Cauchy summation correctly with maple, and maple has different commands for summing up polynomial series. So I use the maple command.

sum((-1)^k*x^(n+k+1)/(factorial(2*k+1)*factorial(n-k)), k = 0 .. n)

Problem is, how can I set a value to n?

Is \sum_{k=0}^n {\frac { \left( -1 \right) ^{k}{x}^{n+k+1}}{ \left( 2\,k+1 \right) !\,\left( n-k \right)!}} the right Cauchy sum of this series anyhow?

 

[Simfish]

test...

The problem here is that this is when a_n = \sin x , b_n = \exp x. When I set b_n = \sin x, a_n = \exp x, I get \sum_{k=0}^n {\frac { \left( -1 \right) ^{n-k}{x}^{n-2k+1}}{ \left( 2\,n - 2k + 1 \right) !\,\left( n)!}}

The other problem is that in the Cauchy formula c_n=\sum_{k=0}^n a_k b_{n-k} I expect a_n and b_n to be symmetrical. But yet when I set u = n - k, where k \in [0,n], u \in [n, 0], which would be a backwards summation. Is there a better way to ensure symmetry of the two terms?