Problem
I need help with simplifying following sum:
$$ 1 + \sum_{i=1}^{\infty}{\frac{1}{i!} * (-1)^i * a * (a + b)^{i-1}} $$
and can get the $a$ out to get
$$ 1 + a*\sum_{i=1}^{\infty}{\frac{1}{i!} * (-1)^i * (a + b)^{i-1}} $$
but really don't know where to go from there. Any help will be appreciated.
Disclamer
If you suspect that this is homework, you are correct. This is part of HW assignment from my linear algebra course. The assignment was to express
$$ exp(A = \begin{pmatrix} -a & b \\ a &-b \end{pmatrix}) $$
in the most simple form. I determined that
$$ A^n = (a+b)^{n-1} * \begin{pmatrix} (-1)^n*a & (-1)^{n-1}*b \\ (-1)^{n-1}*a & (-1)^n*b \end{pmatrix} $$
from that I got the above mentioned sum as value of $exp(A)_{00}$, but don't know how to simplify it further.
As this is a HW, I would appreciate, if you mentioned how you figure out how to simplify it in your answers. Thanks.
Computing $\exp(A)$, where $A=\pmatrix{-a&b\\a&-b}$, is easy once $A$ is diagonalized. To do this, we need to find the eigenvectors of $A$. The characteristic polynomial \begin{align} \chi_A(\lambda) &= \det(A-\lambda E) = \left|\matrix{-a-\lambda & b \\ a & -b-\lambda}\right|= (-a-\lambda)(-b-\lambda)-ab \\&= \lambda^2+(a+b)\lambda \end{align} has roots at $\lambda=0$ and $\lambda=-(a+b)$, so we have $2$ real eigenvalues whenever $a+b\neq 0$. The eigenvectors are $\pmatrix{b\\a}$ and $\pmatrix{-1\\1}$, respectively. Thus, letting $P=\pmatrix{b&-1\\a&1}$ we have $P^{-1} =\frac{1}{a+b}\pmatrix{1&1\\-a&b}$ and $$ P^{-1}AP = \frac{1}{a+b}\pmatrix{1&1\\-a&b} \pmatrix{-a&b\\a&-b} \pmatrix{b&-1\\a&1} = \pmatrix{0&0\\0&-(a+b)}=:D. $$ Now \begin{align} \exp(A) &= \exp(PDP^{-1}) = P\exp(A)P^{-1} = P \pmatrix{1 & 0\\0&e^{-(a+b)}} P^{-1}\\ &= \frac{e^{-(a+b)}}{a+b} \pmatrix{a+be^{a+b}&be^{a+b}-b\\ae^{a+b}-a&ae^{a+b}+b}. \end{align} When $a+b=0$ you have $A^2=0$ so $$ \exp(A) = \sum_{n=0}^\infty \frac{1}{n!} A^n = E + A^1 = \pmatrix{-a+1&b\\a&-b+1}. $$
If you have questions regarding the single steps, just ask and I'll try to explain further.