# Thread: Does this sequence converge?

1. ## Does this sequence converge?

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

2. 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: $\lim _{n \to \infty } \left( { - 1} \right)^{n + 1} \ne 0$.

3. 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?

4. 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 $\left( {a_{2n} } \right) \to -2\;\& \,\left( {a_{2n - 1} } \right) \to 2~?$

5. 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.

6. Originally Posted by Arron
No, thats a bit beyond me.
Well you have a great deal of hard work ahead of you.
Good Luck.