\begin{problem}[10]

\hspace{.2cm}
\begin{problemparts}
\problempart[5]
Give a closed form formula for the product
\[
\prod_{i=1}^{\infty} e^{(2^{-i})}.
\]

\solution[\vspace*{2.5in}]{
\[
\prod_{i=1}^{\infty} e^{ (2^{-i})}
 = \prod_{i=1}^{\infty} \text{exp}(2^{-i})
 =  \text{exp}(\sum_{i=1}^{\infty} 2^{-i})
 =  \text{exp}(1) = e^1
 =  e.
\]
}

\problempart[5] Let
\[
s(n) \eqdef \sum_{k=1}^n\frac{1}{\sqrt{k}}.
\]
The function $s$ is assyptotically equal to $an^b$ for constants $a,b \in
\reals$, \ie $s(n) \sim an^b$.  What is $a$? \brule{0.2in}.  What is $b$?
\brule{0.2in}.

\solution[\vspace*{8cm}]{
$a =2, b = 1/2$.

We can use the integral method to obtain upper and lower bounds on $s(n)$.

\textbf{Upper Bound:}
\[
\sum_{k=1}^n \frac{1}{\sqrt{k}}  \leq  \int_{1}^{n+1}\frac{1}{\sqrt{x}} dx
 =  2 \sqrt{n+1} - 2
 \sim  2 \sqrt{n}
\]


\textbf{Lower Bound:}
\[
\sum_{k=1}^n \frac{1}{\sqrt{k}}  \geq  \int_{1}^{n}\frac{1}{\sqrt{x}} dx
 =  2 \sqrt{n} - 2
 \sim  2\sqrt{n}
\]

Since the sum is bounded above and below by terms asymptotic to
$2\sqrt{n}$, it must itself be asymptotic to $2\sqrt{n}$.
}


\end{problemparts}
\end{problem} 



