# 1. Find the first three terms of the sequence a_n= \frac{(-6)^{sin ((2n+1)^n/2)}}{(n+1)!} 2....

## Question:

1. Find the first three terms of the sequence

{eq}a_n= \frac{(-6)^{sin ((2n+1)^n/2)}}{(n+1)!} {/eq}

2. Find the first three terms of the sequence {eq}a_n= \frac{n!}{(n+4)!} {/eq}

3. {eq}5+.0005+.0000005 +...= {/eq}

4. {eq}\sum_{n=0}^{\infty} \frac{1}{(10n-7)- \frac{1}{(10n+3)}} {/eq}

5. {eq}\sum_{n=0}^{\infty} \frac{-10}{(100n^2-25)}{/eq}

## Definition of Sequence, Geometric Series and Comparison Test:

Let {eq}\displaystyle f:\mathbb{N}\to \mathbb{R} {/eq} be a real valued function, where {eq}\displaystyle \mathbb{N} {/eq} denotes the set of all natural numbers, i.e. the set of all positive integer and {eq}\displaystyle \mathbb{R} {/eq} is set of all real numbers. Then {eq}\displaystyle \{f(n)\}=\{f(1),\;f(2), \;f(3), \cdots \} {/eq} is called a real infinite sequence or simply a sequence. Usually a sequence denoted by {eq}\displaystyle \left\{ {{a_n}} \right\} {/eq}, where {eq}{a_n} = f\left( n \right) ,\; n\in \mathbb{N} {/eq}.

If we have an infinite series in the form {eq}\displaystyle k+kr+kr^2+kr^3+kr^4+ \cdots kr^n
{/eq}, where *k* is a constant and *r* is a common ratio between terms,

then we have an infinite geometric series. When {eq}|r| < 1 {/eq}, the sum of such a series is given by {eq}\displaystyle \frac{k}{1-r} {/eq}.

Sometimes, a series given in summation notation won't be automatically recognizable as a geometric series,

but we can rewrite it using properties of series to take advantage of the geometric series formula.

If {eq}\sum\limits_{u = 1}^\infty {{w_k}} {/eq} and {eq}\sum\limits_{u = 1}^\infty {{c_k}} {/eq} be nonnegative series and

if {eq}\sum\limits_{u = 1}^\infty {{c_k}} {/eq} converges and {eq}0 \leqslant {w_k} \leqslant {c_k} {/eq} for all {eq}k \geqslant u {/eq} ,then {eq}\sum\limits_{l = 1}^\infty {{w_k}} {/eq} converges.

if {eq}\sum\limits_{l = 1}^\infty {{c_k}} {/eq} diverges and {eq}0 \leqslant {c_k} \leqslant {w_k} {/eq} for all {eq}k \geqslant 1 {/eq} ,then {eq}\sum\limits_{l = 1}^\infty {{w_k}} {/eq} diverges.

## Answer and Explanation: 1

{eq}\displaystyle \eqalign{ & 1. \cr & {a_n} = \frac{{{{( - 6)}^{\sin \left( {\frac{{{{(2n + 1)}^n}}}{2}} \right)}}}}{{(n + 1)!}} \cr & {\text{Sequence for the value }}n = 1,2,3. \cr & {a_1} = \frac{{{{( - 6)}^{\sin \left( {\frac{{(2 + 1)}}{2}} \right)}}}}{{(1 + 1)!}} = \frac{{{{( - 6)}^{\sin \left( {\frac{3}{2}} \right)}}}}{2} \cr & {a_2} = \frac{{{{( - 6)}^{\sin \left( {\frac{{{{(4 + 1)}^2}}}{2}} \right)}}}}{{(2 + 1)!}} = \frac{{{{( - 6)}^{\sin \left( {\frac{{25}}{2}} \right)}}}}{6} \cr & {a_3} = \frac{{{{( - 6)}^{\sin \left( {\frac{{{{(6 + 1)}^3}}}{2}} \right)}}}}{{(3 + 1)!}} = \frac{{{{( - 6)}^{\sin \left( {\frac{{343}}{2}} \right)}}}}{{24}} \cr & \cr & 2. \cr & {a_n} = \frac{{n!}}{{(n + 4)!}} \cr & {\text{Sequence for the value }}n = 1,2,3. \cr & {a_1} = \frac{{1!}}{{(1 + 4)!}} = \frac{1}{{120}} \cr & {a_2} = \frac{{2!}}{{(2 + 4)!}} = \frac{1}{{360}} \cr & {a_3} = \frac{{(3!)}}{{7!}} = \frac{1}{{840}} \cr & \cr & 3. \cr & 5 + 0.0005 + 0.0000005 + ........ \cr & 5 + \frac{5}{{{{10}^4}}} + \frac{5}{{{{10}^7}}} + ........ \cr & 5 + 5\left( {\frac{1}{{{{10}^4}}} + \frac{1}{{{{10}^7}}} + ........} \right) \cr & 5 + 5\frac{{\left( {\frac{1}{{{{10}^4}}}} \right)}}{{1 - \left( {\frac{1}{{{{10}^3}}}} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\sum\limits_{n = 0}^\infty {a{r^n} = \frac{a}{{1 - r}},r < 1} } \right) \cr & 5 + 5\frac{{\left( {\frac{1}{{10}}} \right)}}{{1000 - 1}} \cr & 5 + \frac{1}{{1998}} \cr & \frac{{9990 + 1}}{{1998}} \cr & \frac{{9991}}{{1998}} \cr & \cr & 4. \cr & \sum\limits_{n = 0}^\infty {\frac{1}{{(10n - 7) - \frac{1}{{(10n + 3)}}}}} \cr & \sum\limits_{n = 0}^\infty {\frac{{(10n + 3)}}{{(10n - 7)(10n + 3) - 1}}} \cr & {\text{Let,}} \cr & {a_n} = \frac{{(10n + 3)}}{{(10n - 7)(10n + 3) - 1}} \cr & {b_n} = \frac{1}{n} \cr & {a_n} > {b_n},\forall n \cr & {\text{From p - series test }}\sum {{b_n}{\text{ is divergent for }}} p = 1. \cr & {\text{By comparison test }}\sum {{a_n}{\text{ is also divergent}}{\text{.}}} \cr & \cr & 5. \cr & \sum\limits_{n = 0}^\infty {\frac{{ - 10}}{{(100{n^2} - 25)}}} \cr & - 10\sum\limits_{n = 0}^\infty {\frac{1}{{(100{n^2} - 25)}}} \cr & {\text{Let,}} \cr & {a_n} = \frac{1}{{(100{n^2} - 25)}} \cr & {b_n} = \frac{1}{{{n^2}}} \cr & {a_n} < {b_n},\forall n \cr & {\text{From p - series test }}\sum {{b_n}{\text{ is convergent for }}} p = 2 > 1. \cr & {\text{By comparison test }}\sum {{a_n}{\text{ is also convergent}}{\text{.}}} \cr} {/eq}

#### Learn more about this topic:

from

Chapter 28 / Lesson 3Convergence and divergence of a series in math follows some specific rules. Learn the rules as well as the geometric series convergence test. Also see examples.