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

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    $\begingroup$ What's your question? $\endgroup$ – user23793 Nov 29 '18 at 21:52
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    $\begingroup$ The summand is $O(\frac{1}{\sqrt{n}})$, which diverges $\endgroup$ – Alex Nov 29 '18 at 21:53
  • $\begingroup$ My bad. I had the question in my head but did not actually write it out xD $\endgroup$ – Wesley Strik Nov 29 '18 at 21:55
  • $\begingroup$ 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 $\endgroup$ – Wesley Strik Nov 29 '18 at 21:56
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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.

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  • $\begingroup$ That's a nice alternative, we haven't practised a lot with this yet, but this method would work very well :) $\endgroup$ – Wesley Strik Nov 29 '18 at 22:01
  • $\begingroup$ 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. $\endgroup$ – Wesley Strik Nov 29 '18 at 22:02
  • $\begingroup$ Generally if you can't evaluate it directly, but you can "visualize" the series diverges, then this test is good. $\endgroup$ – K Split X Nov 29 '18 at 22:03
  • $\begingroup$ Intuiton can be developed, you're right ;) "seeing it" and a proof are sometimes closer than it seems. $\endgroup$ – Wesley Strik Nov 29 '18 at 22:05
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$$\sum_{n=3}^\infty \frac{\sqrt{n}+2}{n-2}=\sum_{n=3}^\infty \frac1{\sqrt{n}-2}$$ and the terms are of order $n^{-1/2}$.

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Since

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

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

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