Answer
$\sqrt e=1+\dfrac{1}{2}+\dfrac{(\dfrac{1}{2})^2}{2!}+\dfrac{(\dfrac{1}{2})^3}{3!}+.......$
Work Step by Step
The first $4$ non-zero terms of Taylor series can be written as: $e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+.......$
Plug $x=\dfrac{1}{2}$ in the above equation to obtain:
$e^{\dfrac{1}{2}}=1+\dfrac{1}{2}+\dfrac{(\dfrac{1}{2})^2}{2!}+\dfrac{(\dfrac{1}{2})^3}{3!}+.......$
\Sigma
Therefore, the first $4$ non-zero terms of an infinite series is equal to:
$\sqrt e=1+\dfrac{1}{2}+\dfrac{(\dfrac{1}{2})^2}{2!}+\dfrac{(\dfrac{1}{2})^3}{3!}+.......$
