009C Sample Midterm 1, Problem 2

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

ExpandFoundations:
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)

ExpandStep 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
$s_{2}=2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{3}}}{\bigg )}.$

ExpandStep 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
${\begin{array}{rcl}\displaystyle {s_{4}}&=&\displaystyle {2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{3}}}{\bigg )}+2{\bigg (}{\frac {1}{2^{3}}}-{\frac {1}{2^{4}}}{\bigg )}+2{\bigg (}{\frac {1}{2^{4}}}-{\frac {1}{2^{5}}}{\bigg )}}\\&&\\&=&\displaystyle {2{\bigg (}{\frac {1}{2^{2}}}-{\frac {1}{2^{5}}}{\bigg )}.}\end{array}}$

ExpandStep 3:
If we look at $s_{2},s_{3},s_{4},$ 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)

ExpandStep 1:

ExpandStep 2:

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

Return to Sample Exam