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Suppose we know that $a_n \rightarrow 2$, which is a sequence with positive terms, determine whether the following series converges:

$$ \sum \frac{a_n}{n + a_n^2} $$

I would like to apply a variation of the ratio test: $$ \lim_ {n \rightarrow \infty} \frac{\frac{a_{n+1}}{n+1+ a_{n+1}^2}}{\frac{a_n}{n + a_n^2} }= \lim_ {n \rightarrow \infty}\frac{a_{n+1}}{a_n} \left(\frac{n+ a_n^2}{n+1+ a_n^2}\right)=\lim_ {n \rightarrow \infty}\frac{a_{n+1}}{a_n} \left(\frac{1+\frac{a_n^2}{n}}{1+\frac{1}{n}+ \frac{a_{n+1}^2}{n}}\right)= \frac{2}{2} \cdot\frac{1}{1}.$$ We know that $a_n$ converges and it bounded, so we know that $\frac{a_n^2}{n} \rightarrow 0$ But now the ratio test is inconclusive, can someone drop a hint for an alternative approach perhaps?

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  • $\begingroup$ Limit comparison test against $2/n$? $\endgroup$ – Nicholas Stull Nov 28 '18 at 20:09
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This can't converge because the terms of the sequence do not shrink fast enough. Since we know that $a_n\rightarrow 2$, there is some $N$ such that $a_n\in(3/2,5/2)$ for all $n>N$. Then $$ \frac{a_n}{n+a_n^2}>\frac{3/2}{n+25/4}>\frac{1}{n} $$ for all $n>\max(13,N)$. By the comparison test on the tail of your sequence starting from the $\max (13,N)$ we see that your series diverges.

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  • $\begingroup$ I like this argument $\endgroup$ – Wesley Strik Nov 28 '18 at 20:13
  • $\begingroup$ This is basically what we discussed in class, this is very helpful, thank you. $\endgroup$ – Wesley Strik Nov 28 '18 at 20:14
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    $\begingroup$ @WesleyGroupshaveFeelingsToo You are welcome! I think you will find this kind of reasoning will give you a lot of mileage in analysis. Anytime you are given a limit, that also gives you a way of approximating values, and with enough wiggle room (as my professor used to say) you can usually find the proof you want. $\endgroup$ – JDMan4444 Nov 28 '18 at 20:16
  • $\begingroup$ Yes, the trick is always to find a nice approximation, I feel I'm getting more adept at it, but not quite sure when to use this method and when not. I will keep practising. I'm really enjoying analysis so far. $\endgroup$ – Wesley Strik Nov 28 '18 at 20:18
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Hint: Try the sequence $a_n=2$ for all $n$.

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  • $\begingroup$ Cleary $\frac{2}{n+2}$ diverges, because the harmonic series does. I think the question wants an answer for general $a_n$ but this answer inspires the comparison given by other answers. $\endgroup$ – Wesley Strik Nov 28 '18 at 20:16
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Asymptotically this new series $\approx\frac{2}{n+4}$, so diverges.

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By limit comparison test, since eventually $a_n$ is positive

$$\frac{\frac{a_n}{n + a_n^2}}{\frac1n}=\frac{a_n\cdot n}{n + a_n^2}=\frac{a_n}{1 + \frac{a_n^2}n}\to 2$$

therefore the given series diverges.

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Very simple with equivalence:

If $\lim_{n\to\infty}a_n=2$, then $\; a_n\sim_\infty 2$. It is easy to check $\;n+ a_n^2\sim_\infty n$, therefore $$ \frac{a_n}{n+a_n^2}\sim_\infty\frac 2n,$$ which diverges.

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When $a_n\to 2$ there exists $N$ such that $$\forall n>N\qquad,\qquad 1<a_n<3$$therefore $$\sum_{n=1}^{\infty} {a_n\over n+a_n^2}=K+\sum_{n=N}^{\infty} {1\over {n\over a_n}+a_n}\ge K+\sum_{n=N}^{\infty}{1\over {n+3}}=\infty$$therefore the series diverges.

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