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Finding convergence or divergencs of series $$\sum^{\infty}_{n=1}\sqrt{\frac{4n^6+3n}{2n^2+n+5}}$$

What i try::

$$\frac{4n^6+3n}{2n^2+n+5}\approx 2n^4$$

$$\sqrt{\frac{4n^6+3n}{2n^2+n+5}}\approx \sqrt{2}\; n^2$$

$$\sum^{\infty}_{n=1}\sqrt{\frac{4n^6+3n}{2n^2+n+5}}\approx \sqrt{2}\sum^{\infty}_{n=1}n^2$$

Seems that it is divergent.

But i did not justified How do i prove it.Thanks

jacky
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    The $n-$th term is NOT tending to $0$ so divergent. $$\frac{4n^6+3n}{2n^2+n+5} \geq \frac{4n^6}{2n^2+n^2+n^2}=n^4$$ for $n \geq 3$ – Anurag A Jul 05 '20 at 09:57

1 Answers1

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The sucession in your series verifies that: $$\sqrt{\frac{4n^6+3n}{2n^2+n+5}}>0 \text{ (trivial for $n\in\mathbb{N}$),}$$ and $$\lim_{n\rightarrow\infty}\sqrt{\frac{4n^6+3n}{2n^2+n+5}}=\lim_{n\rightarrow\infty}\frac{4n^3}{2n}=\lim_{n\rightarrow\infty}2n^2=\infty.$$ So, given this (positive and going to infinity) we conclude that: $$\sum^{\infty}_{n=1}\sqrt{\frac{4n^6+3n}{2n^2+n+5}}=\infty.$$

Alejandro Bergasa Alonso
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