Determine whether the series  sum_(n=1)^oo (2n^2 +3n)/sqrt(5+n^5) is convergent or divergent. How do i tell which comparison test to use?

Mar 22, 2017

The series:

${\sum}_{n = 1}^{\infty} \frac{2 {n}^{2} + 3 n}{\sqrt{5 + {n}^{5}}}$

is divergent.

Explanation:

We need to determine the convergence of the series:

${\sum}_{n = 1}^{\infty} {a}_{n} = {\sum}_{n = 1}^{\infty} \frac{2 {n}^{2} + 3 n}{\sqrt{5 + {n}^{5}}}$

We can see that the numerator is of order ${n}^{2}$ and the denominator is of order ${n}^{\frac{5}{2}}$.

For $n \to \infty$ then the sequence tends to zero with order ${n}^{- \frac{1}{2}}$ and thus the series will not converge because:

${\sum}_{n = 1}^{\infty} {n}^{- p}$

is convergent only for $p > 1$ (p-series test).

To demonstrate this formally, consider the series:

${\sum}_{n = 1}^{\infty} {b}_{n} = {\sum}_{n = 1}^{\infty} \frac{1}{\sqrt{n}} = {\sum}_{n = 1}^{\infty} \frac{1}{n} ^ \left(\frac{1}{2}\right)$

as stated above this series is divergent based on the p-series test.

Now evaluate the limit:

${\lim}_{n \to \infty} {a}_{n} / {b}_{n} = {\lim}_{n \to \infty} \frac{\frac{2 {n}^{2} + 3 n}{\sqrt{5 + {n}^{5}}}}{\frac{1}{\sqrt{n}}}$

${\lim}_{n \to \infty} {a}_{n} / {b}_{n} = {\lim}_{n \to \infty} \frac{n \sqrt{n} \left(2 n + 3\right)}{\sqrt{5 + {n}^{5}}}$

lim_(n->oo) a_n/b_n = lim_(n->oo) sqrt( (n^3(2n+3)^2)/(5+n^5)

lim_(n->oo) a_n/b_n = lim_(n->oo) sqrt( (n^3(4n^2+12n+9))/(5+n^5)

${\lim}_{n \to \infty} {a}_{n} / {b}_{n} = {\lim}_{n \to \infty} \sqrt{\frac{4 {n}^{5} + 12 {n}^{3} + 9 {n}^{3}}{5 + {n}^{5}}} = 2$

As the limit exists and is finite, based on the limit comparison test also the first series is divergent.