Does this sequence converge?

**Arron** · May 30th 2011, 04:41 AM

Is this correct?

The dominant term is n!, so we write

${a}_{n } = \frac{{n}^{4 }+2(n!) }{{n}^{3 }+{(-1)}^{n+1 }(n!) }$ = $\frac{{n}^{4 }/n! +2 }{{n}^{3 }/n! +{(-1)}^{n+1 } }$

Since { ${n}^{4 }/n!$ },{ ${n}^{3 }/n!$ } and { ${(-1)}^{n+1 }$ } are all null sequences,

$\lim_{n \to \infty} {a}_{n } = \lim_{n \to \infty} \frac{{n}^{4 }/n! +2 }{{n}^{3 }/n! +{(-1)}^{n+1 } }$ = $\frac{0+2}{0+0 }$ = 0

by the Combination Rules.

Hence ${a}_{n }$ converges with a limit 0

**Plato** · May 30th 2011, 04:48 AM

Originally Posted by Arron

Is this correct?
The dominant term is n!, so we write
${a}_{n } = \frac{{n}^{4 }+2(n!) }{{n}^{3 }+{(-1)}^{n+1 }(n!) }$ = $\frac{{n}^{4 }/n! +2 }{{n}^{3 }/n! +{(-1)}^{n+1 } }$
Since { ${n}^{4 }/n!$ },{ ${n}^{3 }/n!$ } and { ${(-1)}^{n+1 }$ } are all null sequences,
$\lim_{n \to \infty} {a}_{n } = \lim_{n \to \infty} \frac{{n}^{4 }/n! +2 }{{n}^{3 }/n! +{(-1)}^{n+1 } }$ = $\frac{0+2}{0+0 }$ = 0
by the Combination Rules.
Hence ${a}_{n }$ converges with a limit 0

Here is a problem with that: $\lim _{n \to \infty } \left( { - 1} \right)^{n + 1} \ne 0$ .

**Arron** · May 30th 2011, 08:25 AM

Are you saying that { ${(-1)}^{n+1 }$ } is not a basic null sequence. I think you are right. Does this mean this is not a converging sequence and how do I prove this?

**Plato** · May 30th 2011, 08:40 AM

Originally Posted by Arron

Are you saying that { ${(-1)}^{n+1 }$ } is not a basic null sequence. I think you are right. Does this mean this is not a converging sequence and how do I prove this?

Can you show that $\left( {a_{2n} } \right) \to -2\;\& \,\left( {a_{2n - 1} } \right) \to 2~?$

**Arron** · May 30th 2011, 09:07 AM

No, thats a bit beyond me.

I am thinking, | ${(-1)}^{n+1 }$ | = 1

Therefore the limit is $\frac{0+2}{0+1 }$ = 2

by the Combination Rules.

**Plato** · May 30th 2011, 09:17 AM

Originally Posted by Arron

No, thats a bit beyond me.

Well you have a great deal of hard work ahead of you.
Good Luck.

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