# 009C Sample Midterm 1, Problem 2 Detailed Solution

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Consider the infinite series  $\sum _{n=2}^{\infty }2{\bigg (}{\frac {1}{2^{n}}}-{\frac {1}{2^{n+1}}}{\bigg )}.$ (a) Find an expression for the  $n$ th partial sum  $s_{n}$ of the series.

(b) Compute  $\lim _{n\rightarrow \infty }s_{n}.$ Background Information:
The  $n$ th partial sum,  $s_{n}$ for a series  $\sum _{n=1}^{\infty }a_{n}$ is defined as

$s_{n}=\sum _{i=1}^{n}a_{i}.$ Solution:

(a)

Step 1:
We need to find a pattern for the partial sums in order to find a formula.
We start by calculating  $s_{2}.$ We have
$s_{2}=2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{3}}}{\bigg )}.$ Step 2:
Next, we calculate  $s_{3}$ and  $s_{4}.$ We have
${\begin{array}{rcl}\displaystyle {s_{3}}&=&\displaystyle {2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{3}}}{\bigg )}+2{\bigg (}{\frac {1}{2^{3}}}-{\frac {1}{2^{4}}}{\bigg )}}\\&&\\&=&\displaystyle {2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{4}}}{\bigg )}}\end{array}}$ and
${\begin{array}{rcl}\displaystyle {s_{4}}&=&\displaystyle {2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{3}}}{\bigg )}+2{\bigg (}{\frac {1}{2^{3}}}-{\frac {1}{2^{4}}}{\bigg )}+2{\bigg (}{\frac {1}{2^{4}}}-{\frac {1}{2^{5}}}{\bigg )}}\\&&\\&=&\displaystyle {2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{5}}}{\bigg )}.}\end{array}}$ Step 3:
If we look at  $s_{2},s_{3},$ and  $s_{4},$ we notice a pattern.
From this pattern, we get the formula
$s_{n}=2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{n+1}}}{\bigg )}.$ (b)

Step 1:
From Part (a), we have
$s_{n}=2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{n+1}}}{\bigg )}.$ Step 2:
We now calculate  $\lim _{n\rightarrow \infty }s_{n}.$ We get
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }s_{n}}&=&\displaystyle {\lim _{n\rightarrow \infty }2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{n+1}}}{\bigg )}}\\&&\\&=&\displaystyle {\frac {2}{2^{2}}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\end{array}}$ (a)     $s_{n}=2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{n+1}}}{\bigg )}$ (b)     ${\frac {1}{2}}$ 