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

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.

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

• What's your question? – user23793 Nov 29 '18 at 21:52
• The summand is $O(\frac{1}{\sqrt{n}})$, which diverges – Alex Nov 29 '18 at 21:53
• My bad. I had the question in my head but did not actually write it out xD – Wesley Strik Nov 29 '18 at 21:55
• Apparently $\frac{1}{\sqrt{n}}$ diverges, I was not sure about this, but with that method I could just carry out the division and conclude this immediately I suppose – Wesley Strik Nov 29 '18 at 21:56

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.

• That's a nice alternative, we haven't practised a lot with this yet, but this method would work very well :) – Wesley Strik Nov 29 '18 at 22:01
• Right now I am still trying to figure out which method would be best suited for which situation, I still feel I'm just throwing all I know at a question most of the time - but I'm getting better. – Wesley Strik Nov 29 '18 at 22:02
• Generally if you can't evaluate it directly, but you can "visualize" the series diverges, then this test is good. – K Split X Nov 29 '18 at 22:03
• Intuiton can be developed, you're right ;) "seeing it" and a proof are sometimes closer than it seems. – Wesley Strik Nov 29 '18 at 22:05

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

Since

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

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