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Section 10.1 : Sequences

4. Determine if the given sequence converges or diverges. If it converges what is its limit?

\[\left\{ {\frac{{{{\left( { - 1} \right)}^{n - 2}}{n^2}}}{{4 + {n^3}}}} \right\}_{n = 0}^\infty \]

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To answer this all we need is the following limit of the sequence terms.

\[\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( { - 1} \right)}^{n - 2}}{n^2}}}{{4 + {n^3}}}\]

However, because of the \({\left( { - 1} \right)^{n - 2}}\) we can’t compute this limit using our knowledge of computing limits from Calculus I.

Show Step 2

Recall however, that we had a nice Fact in the notes from this section that had us computing not the limit above but instead computing the limit of the absolute value of the sequence terms.

\[\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 1} \right)}^{n - 2}}{n^2}}}{{4 + {n^3}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}}}{{4 + {n^3}}} = 0\]

This is a limit that we can compute because the absolute value got rid of the alternating sign, i.e. the \({\left( { - 1} \right)^{n + 2}}\).

Show Step 3

Now, by the Fact from class we know that because the limit of the absolute value of the sequence terms was zero (and recall that to use that fact the limit MUST be zero!) we also know the following limit.

\[\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( { - 1} \right)}^{n - 2}}{n^2}}}{{4 + {n^3}}} = 0\] Show Step 4

We can see that the limit of the terms existed and was a finite number and so we know that the sequence converges and its limit is zero.