Question

(1 point) by recognizing each series below as a taylor series evaluated at a particular value of x, find the sum of each convergent series.
a. 1+2+222!+233!+244!+⋯+2nn!+⋯=
b. 1−422!+444!−466!+⋯+(−1)n42n(2n)!+⋯=

Answer 1

Guessing on what you're trying to say:

(a)
$1+2+\dfrac{2^2}{2!}+\dfrac{2^3}{3!}+\cdots+\dfrac{2^n}{n!}+\cdots=\displaystyle\sum_{n=0}^\infty\frac{2^n}{n!}$

Compare to the series

$e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}$

which means the value of the sum above is $e^2$.

(b)
$1-\dfrac{4^2}{2!}+\dfrac{4^4}{4!}-\cdots+\dfrac{(-1)^n4^{2n}}{(2n)!}+\cdots=\displaystyle\sum_{n=0}^\infty\frac{(-1)^n4^{2n}}{(2n)!}$

Compare to

$\cos x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}$

so that the sum evaluates to $\cos4$.