# A) Show series converges or diverges using the comparison test: \sum_{n=1}^\infty \frac{6n +...

## Question:

A) Show series converges or diverges using the comparison test:

{eq}\displaystyle \sum_{n=1}^\infty \frac{6n + 5}{n(n+1)(n + 2)}{/eq}

B) Show series converges or diverges using the ration test:

{eq}\displaystyle \sum_{n = !}^ \infty \frac{(n + 2)(n + 8)}{n!}{/eq}

## Series Convergence:

An infinite series {eq}\sum\limits_{n = 1}^\infty {{u_n}} {/eq} is said to be convergent if the sequence {eq}{S_n} = \sum\limits_{p = 1}^n {{u_p}} {/eq} of an nth partial sum is convergent.

Comparison test is used when we are able to compare the given series to one whose convergence properties are known.

Limit comparision Test:

Let {eq}\sum\limits_{n = 1}^\infty {{u_n}} ,\sum\limits_{n = 1}^\infty {{v_n}} {/eq} be two infinite series of positive terms such that,

{eq}\mathop {\lim }\limits_{n \to \infty } \frac{{{u_n}}}{{{v_n}}} = c {/eq} where {eq}c {/eq} is finite, positive real number.

then either both series converge or both series diverge.

Ratio Test:

Let {eq}\sum\limits_{n = 1}^\infty {{u_n}} {/eq} be an infinite series of positive terms and {eq}\mathop {\lim }\limits_{n \to \infty } \frac{{{u_{n + 1}}}}{{{u_n}}} = L {/eq}

then the following hold:

{eq}\eqalign{ & 1) \ \ L > 1 \Rightarrow \text{the series is divergent} \cr & 2) \ \ L < 1 \Rightarrow \text{the series is convergent} \cr & 3) \ \ L = 1 \Rightarrow \text{the test is inconclusive} \cr} {/eq}

{eq}\displaystyle \text{The infinite series}~\sum\limits_{n = 1}^\infty {\frac{1}{{{n^p}}}} ~\text{is convergent if }~p>1 {/eq}

## Answer and Explanation:

We have,

{eq}\displaystyle \sum_{n=1}^\infty \frac{6n + 5}{n(n+1)(n + 2)} {/eq}

Let us apply limit comparision test,

{eq}\eqalign{ \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{6n + 5}}{{n(n + 1)(n + 2)}}}}{{\frac{1}{{{n^2}}}}} &= \mathop {\lim }\limits_{n \to \infty } \frac{{6n + 5}}{{n(n + 1)(n + 2)}}\frac{{{n^2}}}{1} \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^3}\left( {6 + \frac{5}{n}} \right)}}{{{n^3}\left( {1 + \frac{1}{n}} \right)\left( {1 + \frac{2}{n}} \right)}} \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {6 + \frac{5}{n}} \right)}}{{\left( {1 + \frac{1}{n}} \right)\left( {1 + \frac{2}{n}} \right)}} \cr & = \frac{{\left( {6 + \frac{5}{\infty }} \right)}}{{\left( {1 + \frac{1}{\infty }} \right)\left( {1 + \frac{2}{\infty }} \right)}} \cr & = 6 \cr} {/eq}

We have, the 2-series {eq}\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} {/eq} convergent, hence by limit comparison test the series {eq}\sum_{n=1}^\infty \frac{6n + 5}{n(n + 1)(n + 2)} {/eq} is convergent.

B) Show series converges or diverges using the ration test:

{eq}\displaystyle \sum_{n = !}^ \infty \frac{(n + 2)(n + 8)}{n!} {/eq}

Let us apply the ratio test,

{eq}\eqalign{ \mathop {\lim }\limits_{n \to \infty } \frac{{{u_{n + 1}}}}{{{u_n}}} &= \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{(n + 1 + 2)(n + 1 + 8)}}{{\left( {n + 1} \right)!}}}}{{\frac{{(n + 2)(n + 8)}}{{n!}}}} \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{{(n + 3)(n + 9)}}{{\left( {n + 1} \right)!}}\frac{{n!}}{{(n + 2)(n + 8)}} \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}\left( {1 + \frac{3}{n}} \right)\left( {1 + \frac{9}{n}} \right)}}{{\left( {n + 1} \right).n!}}\frac{{n!}}{{{n^2}\left( {1 + \frac{2}{n}} \right)\left( {1 + \frac{8}{n}} \right)}} \cr & = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {1 + \frac{3}{n}} \right)\left( {1 + \frac{9}{n}} \right)}}{{\left( {n + 1} \right)}}\frac{1}{{\left( {1 + \frac{2}{n}} \right)\left( {1 + \frac{8}{n}} \right)}} \cr & = \frac{{\left( {1 + \frac{3}{\infty }} \right)\left( {1 + \frac{9}{\infty }} \right)}}{{\left( {\infty + 1} \right)}}\frac{1}{{\left( {1 + \frac{2}{\infty }} \right)\left( {1 + \frac{8}{\infty }} \right)}} \cr & = \frac{{\left( {1 + 0} \right)\left( {1 + 0} \right)}}{{\left( {\infty + 1} \right)}}\frac{1}{{\left( {1 + 0} \right)\left( {1 + 0} \right)}}~~~~~\left( {\because~\text{ for any finite number}~\frac{a}{\infty } = 0} \right) \cr & = \frac{1}{\infty } \cr & = 0 <1\cr} {/eq}

Hence by ratio test series {eq}\displaystyle \sum_{n = !}^ \infty \frac{(n + 2)(n + 8)}{n!}{/eq} is convergent.