009C Sample Midterm 1, Problem 2 Detailed Solution

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  $\lim _{n\rightarrow \infty }s_{n}.$

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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
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }s_{n}}&=&\displaystyle {\lim _{n\rightarrow \infty }2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{n+1}}}{\bigg )}}\\&&\\&=&\displaystyle {\frac {2}{2^{2}}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\end{array}}$

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

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