Determine if $\sum\limits_{n=0}^{\infty}\left(\frac{3n^2+1}{n^2+2}\right)^{5n}$ converges. Determine if $$\sum_{n=0}^{\infty}\left(\frac{3n^2+1}{n^2+2}\right)^{5n}$$ converges.
My question is rather strange. At first I thought of applying the divergence test. We see that as $n\to\infty$, the inside term approaches $3$, and then we have $3^\infty$, which is clearly $\infty$. So I simply used divergence test to solve this.
When I entered my answer in a couple of calculators online to check my answer, they did not use divergence test, but used ratio test. Usually if something is as easy as the divergence test, they use it first, but they did not.
This made me think, is my limit evaluation technique incorrect? Can I not say that the sum approaches $3^\infty$? Is there flawed logic here?
 A: $$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{\left(\frac{3\left(n+1\right)^2+1}{\left(n+1\right)^2+2}\right)^{5\left(n+1\right)}}{\left(\frac{3n^2+1}{n^2+2}\right)^{5n}}\right|\rightarrow\lim _{n\to \infty }\left(\left|\frac{\left(3n^2+6n+4\right)^{5\left(n+1\right)}\left(n^2+2\right)^{5n}}{\left(n^2+2n+3\right)^{5\left(n+1\right)}\left(3n^2+1\right)^{5n}}\right|\right) = \color{red}{243}$$
A: The ratio test can be helpful for an infinite series that "looks like" $\sum 0^{\infty}$. 
For example, it can be used to show that $\sum_{n \geq 1}(1/n)^n$ converges.
Of course, if a sequence does not converge to zero, then its corresponding series will diverge; so, probably someone could alter the algorithm in these online calculators to check, first, whether the parenthetical portion does converge to zero (and, in your case, would find that, no, it converges to three; so, the series diverges). 
But I suspect many series that are put into these calculators are not quite so simply resolved, so that this extra check would generally be inconclusive, hence add to the computation time without settling the matter; meanwhile, my guess is that whatever online calculator you used for this particular series (e.g., WolframAlpha) carried out the ratio test to determine divergence quickly.
A: For $n\geq 0$, the function $n^2+2$ is  increasing and positive  so $1/(n^2+2)$ is  decreasing so $5/(n^2+2)$ is decreasing so $-5/(n^2+2)$ is increasing so $$F(n)=3-5/(n^2+2)=(3n^2+1)/(n^2+2)$$ is increasing. And $F(1)=4/3$.So for $n\geq 1$ we have $$F(n)^{5n}\geq F(1)^{5n}=(4/3)^{5n}>4/3.$$ As the individual terms do not converge to $0$, the series diverges.
Instead of saying anything about $3^{\infty}$, you could say that, as $F(n)\to 3$ as $n\to \infty ,$ we have $F(n)>5/2$ for all but finitely many $n$, so $F(n)^{5n}>(5/2)^{5n}$ for all but finitely many $n$, so the terms form an unbounded sequence, which of course means the series is not summable.  
A: You can certainly use the divergence test, but one may need to be more careful when doing so, since the limit of the $n^{\rm th}$ summand may be an indeterminate form.  In your case, it is not, since for $n$ sufficiently large (e.g. $n \ge 2$), we can bound the summand below as follows:  $$a_n = \left(\frac{3n^2+1}{n^2+2}\right)^{5n} = \left( 3 - \frac{5}{n^2+2} \right)^{5n} > (3-1)^{5n} > 2^5 > 0.$$  This is sufficient to prove that the infinite sum is divergent.
Alternatively, we can show that $\lim_{n \to \infty} \log a_n > -\infty$; i.e., $$\lim_{n \to \infty} 5n \log \frac{3n^2+1}{n^2+2} = \lim_{n \to \infty} 5n \log \left(\frac{3 + 1/n^2}{1 + 2/n^2}\right) = 5 \lim_{n \to \infty} n \lim_{n \to \infty} \log \frac{3 + 1/n^2}{1 + 2/n^2} = 5 \log 3 \lim_{n \to \infty} n = \infty,$$ so in fact even the logarithm of the summand is unbounded: meaning that not only does the summand not tend to $0$, and not only is it increasing, but it increases exponentially without bound, and the sequence is not even convergent, never mind the infinite series.
A: Just show $3n^2 + 1 > n^2 +2, n=1,2,\dots.$ This implies the $n$th term of the series is $> 1$ for $n\ge 1.$ Thus the terms of the series do not $\to 0,$ and we're done.
