# Determine the convergence or divergence of the series. sum_{n=1}^inftyfrac{n}{sqrt{3n^2+3}}

Alyce Wilkinson 2020-11-08 Answered
Determine the convergence or divergence of the series.
$\sum _{n=1}^{\mathrm{\infty }}\frac{n}{\sqrt{3{n}^{2}+3}}$
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## Expert Answer

averes8
Answered 2020-11-09 Author has 92 answers
Consider the given series $\sum _{n=1}^{\mathrm{\infty }}\frac{n}{\sqrt{3{n}^{2}+3}}$
Divergence test series,
If $\underset{n\to \mathrm{\infty }}{lim}{a}_{n}\ne 0$ then series $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}$ is diverge
And
If $\underset{n\to \mathrm{\infty }}{lim}{a}_{n}=0$ then series $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}$ is is convergence or diverge
Given ${a}_{n}=\frac{n}{\sqrt{3{n}^{2}+3}}$
Now find the limit
$\underset{n\to \mathrm{\infty }}{lim}{a}_{n}=\underset{n\to \mathrm{\infty }}{lim}\frac{n}{\sqrt{3{n}^{2}+3}}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{n}{n\sqrt{3+\frac{3}{{n}^{2}}}}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{n}{n\sqrt{3+\frac{3}{\left(\mathrm{\infty }{\right)}^{2}}}}$
$=\frac{1}{\sqrt{3}}$
Hence, the series is divergence because $\underset{n\to \mathrm{\infty }}{lim}{a}_{n}\ne 0$
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