"Determine the convergence or divergence of the series...." was answered by our community

Alyce Wilkinson

Answered question

2020-11-08

Determine the convergence or divergence of the series.

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

averes8

Skilled2020-11-09Added 92 answers

Consider the given series

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

Divergence test series,
If

lim_{n \to \infty} a_{n} \neq 0

then series

\sum_{n = 1}^{\infty} a_{n}

is diverge
And
If

lim_{n \to \infty} a_{n} = 0

then series

\sum_{n = 1}^{\infty} a_{n}

is is convergence or diverge
Given

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

Now find the limit

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

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

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

= \frac{1}{\sqrt{3}}

Hence, the series is divergence because

lim_{n \to \infty} a_{n} \neq 0

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?