"Use the Direct Comparison Test or the Limit..." was answered by our community

Lipossig

Answered question

2020-12-28

Use the Direct Comparison Test or the Limit Comparison Test to determine the convergence or divergence of the series.

\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^{3} + 2 n}}

au4gsf

Skilled2020-12-29Added 95 answers

Limit comparison test:
Suppose that we have two series

\sum a_{n}

and

\sum b_{n}

, define,

b_{n} = \frac{1}{n^{\frac{3}{2}}}

Where L is positive and finite then either both series are convergent or both series divergent.
Given that,

\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^{3} + 2 n}}

Since we have,

\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^{3} + 2 n}}

Here

a_{n} = \frac{1}{\sqrt{n^{3} + 2 n}}

So,

b_{n} = \frac{1}{n^{\frac{3}{2}}}

Now evaluate

b_{n} = \frac{1}{n^{\frac{3}{2}}}

lim_{n \to \infty} \frac{a_{n}}{b_{n}} = lim_{n \to \infty} \frac{\frac{1}{\sqrt{n^{3} + 2 n}}}{\frac{1}{n^{\frac{3}{2}}}}

= lim_{n \to \infty} (\frac{1}{\sqrt{n^{3} + 2 n}} \times n^{\frac{3}{2}})

= lim_{n \to \infty} (\frac{1}{n^{\frac{3}{2}} \sqrt{1 + \frac{2}{n^{2}}}} \times n^{\frac{3}{2}})

= lim_{n \to \infty} (\frac{1}{\sqrt{1 + \frac{2}{n^{2}}}})

= 1

Since

lim_{n \to \infty} \frac{a_{n}}{b_{n}} = L = 1

which is finite and positive.
Since using p-series test [

\sum \frac{1}{n^{p}}

is convergent if

p > 1

or divergent if

p \leq 1

b_{n} = \frac{1}{n^{\frac{3}{2}}}

is convergent.
Since

lim_{n \to \infty} \frac{a_{n}}{b_{n}}

is finite and positive and

\sum b_{n}

is convergent so using limit comparison test

\sum a_{n}

is also convergent.

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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