Question

question 3 of 13
use the limit comparison test for
\\(\sum_{n = 0}^{\infty}a_{n}=\sum_{n = 0}^{\infty}\frac{2n^{2}+15}{n(n - 4)(n - 5)}\\)
to prove convergence or divergence of the infinite series.
(use symbolic notation and fractions where needed.)

\\(b_{n}=\\)

\\(\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=\\)

basing on the obtained value, conclude that
\\(\circ\\) the series converges.
\\(\circ\\) the series diverges.

Explanation:

Step1: Find a suitable \(b_n\)

For large \(n\), the dominant terms in \(a_n=\frac{2n^2 + 15}{n(n - 4)(n - 5)}\) are considered. Expand the denominator \(n(n - 4)(n - 5)=n(n^2-9n + 20)=n^3-9n^2+20n\). The leading - term of the numerator is \(2n^2\) and of the denominator is \(n^3\). So, we choose \(b_n=\frac{1}{n}\) (since \(\frac{2n^2}{n^3}=\frac{2}{n}\)).

Step2: Calculate \(\lim_{n

ightarrow\infty}\frac{a_n}{b_n}\)

$$
LATEXBLOCK0
$$

As \(n
ightarrow\infty\), \(\frac{15}{n^2}
ightarrow0\), \(\frac{9}{n}
ightarrow0\) and \(\frac{20}{n^2}
ightarrow0\). So, \(\lim_{n
ightarrow\infty}\frac{a_n}{b_n}=2\).

Step3: Determine convergence or divergence

The series \(\sum_{n = 1}^{\infty}b_n=\sum_{n = 1}^{\infty}\frac{1}{n}\) is the harmonic series, which diverges. Since \(\lim_{n
ightarrow\infty}\frac{a_n}{b_n}=2>0\), by the Limit - Comparison Test, the series \(\sum_{n = 1}^{\infty}a_n\) diverges.

Answer:

\(b_n=\frac{1}{n}\), \(\lim_{n
ightarrow\infty}\frac{a_n}{b_n}=2\), the series diverges.