\begin{problem}[15]

\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}]{
\begin{eqnarray*}
\prod_{i=1}^{\infty} e^{ (2^{-i})} 
& = & e^{\sum_{i=1}^{\infty} 2^{-i}} \\
& = & e^1\\ 
& = & e	
\end{eqnarray*}
}

\instatements{\newpage}

\problempart[10] Let
\[
s(n) \eqdef \sum_{k=1}^n\frac{1}{\sqrt{k}}.
\]
Use the integral method to obtain upper and lower bounds on $s(n)$ that
enable you to conclude that $s(n) \sim 2\sqrt{n}$.

\solution{
\begin{eqnarray*}
\sum_{k=1}^n \frac{1}{\sqrt{k}} & \leq & \int_{0}^{n}\frac{1}{\sqrt{x}} dx\\
& = & 2 \sqrt{n}\\
& \sim & 2 \sqrt{n}
\end{eqnarray*}

\begin{eqnarray*}
\sum_{k=1}^n \frac{1}{\sqrt{k}} & \geq & \int_{1}^{n+1}\frac{1}{\sqrt{x}} dx\\
& = & 2 (\sqrt{n+1} - 1)\\
& \sim & 2\sqrt{n}
\end{eqnarray*}

It follows that $\sum_{k=1}^n \frac{1}{\sqrt{k}} = \Theta(2 \sqrt{n})$.
}

\iffalse
\problempart 
Find a function $g(n)$ such that $\sum_{k=1}^{n} k^5=\Theta(g)$. 

\solution{
\begin{eqnarray*}
\sum_{k=1}^{n} k^5 
&\leq& \int_{1}^{n+1} x^5 dx	\\
&  = & \frac{(n+1)^6 -1}{6}	\\
&\sim& n^6/6
\end{eqnarray*}

\begin{eqnarray*}
\sum_{k=1}^{n} k^5 & \geq & \int_{0}^{n} x^5 dx\\
& = & \frac{n^6}{6}\\
& \sim & n^6/6
\end{eqnarray*}

It follows that $\sum_{k=1}^{n} k^5 = \Theta(n^6/6)$.
}
\fi


\end{problemparts}
\end{problem} 

\comment{Radhika}{Albert and I discussed alternatively wording it so
 that $\sum_{k=1}^n \frac{1}{\sqrt{k}} \approxeq c.g(n)$ for a certain
 constant $c$. But I'm not very sure about the right way to put
 it. Check with Albert on this one.}

