Let $f(x) = \sum_{ n=0 }^{ \infty } c_n(x-a)^n $ be a power series centered at $a$ with radius of convergence $0< R< \infty$. If the power series converges at $a+R$, then $f$ is continuous at $a+R$, i.e.
$$\limd{ x }{ a+R }{ x \in (a-R, a+R)} \sum_{ n=0 }^{ \infty }c_n(x-a)^n = \sum_{ n=0 }^{ \infty }c_n R^n $$
Similarly, if the power series converges at $a-R $, then $f$ is continuous at $a-R $, i.e.
$$\limd{ x }{ a-R }{ x \in (a-R, a+R)} \sum_{ n=0 }^{ \infty }c_n(x-a)^n = \sum_{ n=0 }^{ \infty }c_n {(-R)}^n $$
Let $(a^{(n)})^\infty_{ n =0} \in \R, (b^{(n)})^\infty_{ n =0} \in \R. \limu{ n }{ \infty }a^{(n)} = A, \limu{ n }{ \infty }b^{(n)} = B$.
Suppose that the sum $\sum_{ n=0 }^{ \infty } (a_{n+1} - a_n) b_n $ is convergent. Then the sum $\sum_{ n=0 }^{ \infty }a_{n+1}(b_{n+1} - b_n) $ is also convergent, and
$$\sum_{ n=0 }^{ \infty }(a_{n+1} - a_n) b_n = AB - a_0b_0 - \sum_{ n=0 }^{ \infty }a_{n+1}(b_{n+1} -b_n)$$
One should compare this formula with the more well-known integration by parts formula
$$\int_{ 0 }^{ \infty } f'(x)g(x) \d x = f(x)g(x)|^ \infty_0 - \int_{ 0 }^{ \infty } f(x)g'(x) \d x $$