Question

does the sequence \\( \\{ a _ { n } \\} \\) converge or diverge? find the limit if the sequence is convergent. \\( a _ { n } = \frac { 1 + 4 n ^ { 4 } } { n ^ { 4 } + 2 n ^ { 3 } } \\)

Explanation:

Step1: Divide numerator and denominator by \(n^{4}\)

$$\begin{align*} a_{n}&=\frac{1 + 4n^{4}}{n^{4}+2n^{3}}\\ &=\frac{\frac{1}{n^{4}}+\frac{4n^{4}}{n^{4}}}{\frac{n^{4}}{n^{4}}+\frac{2n^{3}}{n^{4}}}\\ &=\frac{\frac{1}{n^{4}} + 4}{1+\frac{2}{n}} \end{align*}$$

Step2: Find the limit as \(n

ightarrow\infty\)
We know that \(\lim_{n
ightarrow\infty}\frac{1}{n^{k}} = 0\) for \(k>0\).
So, \(\lim_{n
ightarrow\infty}a_{n}=\lim_{n
ightarrow\infty}\frac{\frac{1}{n^{4}} + 4}{1+\frac{2}{n}}\)
Substitute \(\lim_{n
ightarrow\infty}\frac{1}{n^{4}}=0\) and \(\lim_{n
ightarrow\infty}\frac{2}{n}=0\)
\(\lim_{n
ightarrow\infty}a_{n}=\frac{0 + 4}{1+0}\)

Answer:

The sequence \(\{a_{n}\}\) converges and \(\lim_{n
ightarrow\infty}a_{n}=4\)