divergence of $\sum_{n=3}^\infty \frac{\sqrt{n}+2}{n-2}$ verification/ alternative method

Question

I wish to prove divergence of $$\sum_{n=3}^\infty \frac{\sqrt{n}+2}{n-2}$$

I wish to do so by comparison, since $n\geq 3$: $$\sum_{n=3}^\infty \frac{\sqrt{n}+2}{n-2} > \sum_{n=3}^\infty \frac{1+2}{n-2}>\sum_{n=3}^\infty \frac{3}{n}>\sum_{n=3}^\infty \frac{1}{n} \rightarrow \infty$$ And the harmonic series is divergent, so if we just remove finitely many terms, we still have that it is divergent, because divergence is determined "in the tail". We have a divergent minorant series and hence the original series diverges to $\infty$.

Is this approach fine, or is there some more elegant method, this was about the simplest thing I could think of.

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Alternatively we have: $$\sum_{n=3}^\infty \frac{\sqrt{n}+2}{n-2} > \sum_{n=3}^\infty \frac{\sqrt{n}+2}{n}=\sum_{n=3}^\infty \frac{1}{\sqrt{n}}+ \frac{2}{n}\rightarrow \infty$$

Answer 1

Not clear what your question is, but your answer is correct.

Your approach is fine. Comparison test would be the proper test to use.

One can also show that $$\sum _{n=3}^{\infty \:}\frac{\sqrt{n}+2}{n-2}\ge \sum _{n=3}^{\infty \:}\frac{\sqrt{n}+2}{n}$$

and show that the rightmost sum is diverging via the integral test.

Answer 2

Since

$$\frac{\sqrt{n}+2}{n-2} \sim \frac {\sqrt n}n=\frac1{\sqrt n}$$

the series diverges by limit comparison test with $\sum \frac1{\sqrt n}$.

Answer 3

$$\sum_{n=3}^\infty \frac{\sqrt{n}+2}{n-2}=\sum_{n=3}^\infty \frac1{\sqrt{n}-2}$$ and the terms are of order $n^{-1/2}$.