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%Fall 2000 Quiz 2
% Added by George
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\problem\pts{10} \textbf {Bounding Infinite Sums}
% not multiple choice (but could be)

Using the Integral Method, give an upper bound for 
\[\sum_{i=1}^n \frac{3}{i^{10}} \] 
 
\solution{

\textbf{Solution:}\\
\medskip
Since $f(k)=\frac{3}{i^{10}}$ is monotonically decreasing for $x>1$, we have:
    \begin{eqnarray*}
      \sum_{i=1}^n{\frac{3}{i^{10}}} &=& 3 + \sum_{i=2}^n{\frac{3}{i^{10}}} \\
      &\leq& 3+ \int_1^n{\frac{3}{x^{10}}}dx\\
      &=& 3+ 3\int_1^n{\frac{1}{x^{10}}}dx\\
      &=& 3 + 3\left [{\frac{-1}{9x^{9}}}\right ]_1^n\\
      &=& 3 + \left [{\frac{-1}{3x^{9}}}\right ]_1^n\\
      &=& 3 - \frac{1}{3n^{9}} + \frac{1}{3}\\
      &=& \frac{10}{3} - \frac{1}{3n^{9}}
    \end{eqnarray*}
    Hence, $\sum_{i=1}^n \frac{3}{x^{10}} \leq \frac{10}{3} - \frac{1}{3n^{9}}$.

}


 
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%Fall 2000 Quiz 2
% Added by George
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\problem\pts{25} \textbf {Summations}

Find the closed form for each of the expressions listed below.  Give
your answer in simplified form.

\problempart {\bf (10 points)}: $ \sum_{i=x}^{y} 2i+1 $

\solution{ \bigskip
{\bf Solution:}

There are two ways of doing this problem.  In the first, we re-form the 
sum so the lower limit is 0 instead of $x$ and apply the arithmetic sum 
formula.  

$$\sum_{i=x}^y 2i+1 = \sum_{i=0}^{y-x} 2(i+x)+1$$
$$= \sum_{i=0}^{y-x} 2i + 2x+1$$
$$= (y-x+1)(2x+1) + 2 \sum_{i=0}^{y-x} i $$
$$= (y-x+1)(2x+1) + 2 \frac{(y-x)(y-x+1)}{2} $$
$$= (y-x+1)(2x+1 + y-x) = (y-x+1)(y+x+1) = ((y+1)-x)((y+1)+x) $$
$$= (y+1)^2 - x^2.$$


In the second, we express this sum as the difference of two summations,
one involving only $y$ and one involving only $x$.  

$$\sum_{i=x}^{y} 2i+1 = \sum_{i=0}^{y} 2i + 1 - (\sum_{0=1}^{x-1} 2i +
1).$$

Now, since 

$$\sum_{i=0}^n 2i+1 = n+1 + 2\frac{n(n+1)}{2} = n+1 + n(n+1) = n^2 +
2n+1 = (n+1)^2,$$

we can evaluate this to

$$(y+1)^2 - x^2.$$
}
\mediumspace
\problempart {\bf (15 points)}:
$ \sum_{i=0}^{\infty} \sum_{j=1}^n \left(\frac{j}{j+2}\right)^i $

\solution{ \bigskip
{\bf Solution:}

First, we reorder the summations to get

$$\sum_{j=1}^n \sum_{i=0}^{\infty} \left(\frac{j}{j+2}\right)^i.$$

We can do this reordering because the value of $j$ never depends on the
value of $i$, we just use every possible pair of values for a nonnegative 
$i$ and a $j$ between 1 and $n$.

Evaluating the inner summation, which is just an infinite geometric
series, we get

$$\sum_{j=1}^n \frac{1}{1 - \frac{j}{j+2}}$$
$$= \sum_{j=1}^n \frac{j+2}{2}$$
$$= n + \frac{1}{2}\sum_{j=1}^n j$$
$$= n + \frac{n(n+1)}{4}$$

which simplifies to $\frac{1}{4}n^2 + \frac{5}{4}n$.

}

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%Fall 1998 Midterm
% Added by George
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\problem {\bf (5 points)} Find a closed-form expression equal to the
following summation.

\[
\sum_{j=1}^n \sum_{i=0}^{\infty} \sqrt{j} \cdot
    \left( 1 - \frac{1}{2\sqrt{j}} \right)^i
\]

\solution{
\begin{eqnarray*}
\sum_{j=1}^n
    \sum_{i=0}^{\infty}
        \sqrt{j} \cdot \left( 1 - \frac{1}{2\sqrt{j}} \right)^i
    & = &
\sum_{j=1}^n \frac{\sqrt{j}}{1 - \left( 1 - \frac{1}{2\sqrt{j}} \right)} \\
    & = &   2 \cdot \sum_{j=1}^n j \\
    & = &   2 \cdot \frac{n (n+1)}{2} \\
    & = &   n (n + 1)
\end{eqnarray*}

First, we use the formula for the sum of an infinite geometric series
to eliminate the inner summation.  After simplifying, we apply the
formula for the sum of an arithmetic series.}

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%Fall 2000 Final exam
% Added by George
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%% PROBLEM 5
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\problem\pts{6} \textbf {Summations}

\problempart {\bf (2 points)}  Find a simplified closed form for

$$\sum_{i=0}^{\infty} \sum_{j=1}^n \frac{1}{2^i}$$

\solution{\bigskip {\bf Solution:}

The inner sum, 

$$ \sum_{j=1}^n \frac{1}{2^i} $$

is very simple since $j$ is not involved in the inner term.  The
closed form for this is simply

$$ \frac{n}{2^i} $$.

Since $n$ is a constant, we can take it out, and evaluate a simple
infinite geometric sum to get $\sum_{i=0}^{\infty} \sum_{j = 1}^n
\frac{1}{2^i} = 2n$
}

\mediumspace

\problempart {\bf (4 points)}  Find a simplified closed form for

$$\sum_{i=0}^n (i+1)x^i.$$

\solution{\bigskip {\bf Solution:}

Consider that 

$$\frac{d}{dx}\left( \sum_{i=0}^{n+1} x^i \right) = \sum_{i=0}^n (i+1)x^i.$$

However, the sum $\sum_{i=0}^{n+1} x^i$ can be written in closed form as

$$\frac{x^{n+2} - 1}{x - 1}.$$

Taking the derivative of this yields

$$\frac{(n+2)x^{n+1}}{x-1} - \frac{x^{n+2} - 1}{(x-1)^2} = $$

$$\frac{(n+1)x^{n+2} - (n+2)x^{n+1} + 1}{(x-1)^2}.$$
}
