# 009C Sample Midterm 1, Problem 2 Detailed Solution

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Consider the infinite series  ${\displaystyle \sum _{n=2}^{\infty }2{\bigg (}{\frac {1}{2^{n}}}-{\frac {1}{2^{n+1}}}{\bigg )}.}$

(a) Find an expression for the  ${\displaystyle n}$th partial sum  ${\displaystyle s_{n}}$  of the series.

(b) Compute  ${\displaystyle \lim _{n\rightarrow \infty }s_{n}.}$

Background Information:
The  ${\displaystyle n}$th partial sum,  ${\displaystyle s_{n}}$  for a series  ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$  is defined as

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

Solution:

(a)

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

(b)

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

(a)     ${\displaystyle s_{n}=2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{n+1}}}{\bigg )}}$
(b)     ${\displaystyle {\frac {1}{2}}}$