Difference between revisions of "009C Sample Midterm 1, Problem 2"

Revision as of 13:59, 12 February 2017

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}.

Foundations:
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 $s_{2},s_{3},s_{4},$ 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)

Final Answer:
(a)
(b)

@@ Line 34: / Line 34: @@
 |Next, we calculate <math>s_3</math> and <math>s_4.</math> We have
 |-
-|
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\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}</math>
+|-
+|and
+|-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\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}</math>
+|}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+!Step 3: &nbsp;
+|-
+|If we look at <math>s_2,s_3,s_4, </math> we notice a pattern.
+|-
+|From this pattern, we get the formula
+|-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg).</math>
 |}