2019
Problem - 4573
Let $f_0(x)=(\sqrt{e})^x$ , and recursively define $f_{n+1}(x) = f'_n(x)$ for integers $n\ge 0$. Compute $$\sum_{k=0}^{\infty}f_k(1)$$
Rewrite $f_0(x)=e^{\frac{x}{2}}$. Then $f_1(x)=\frac{1}{2}e^{\frac{x}{2}}$. By induction, we can find that $f_n(x)=\frac{1}{2^n}e^{\frac{x}{2}}$. Hence
$$\sum_{k=0}^{\infty}f(1)=\sum_{k=0}^{\infty}\frac{1}{2^k}e^{\frac{1}{2}}=\sqrt{e}\sum_{k=0}^{\infty}\frac{1}{2^k}=\boxed{2\sqrt{e}}$$