Root Test Using Ln()? Root test for $$A_n:\sum\frac n{n^2+1}.$$
When we take limit of the function, I know that $n^{1/n}$ is going to $1$.
But what happens to the denominator? As I understand, an $\ln()$ trick is used.
 A: The root test says to examine the value of $\displaystyle\lim_{n\to+\infty} |A_n|^{1/n}$, where in this case, $A_n = \dfrac n{n^2+1}$.
Let's use the fact that $$ \lim_{n\to+\infty} |A_n|^{1/n} = \lim_{x\to+\infty} |f(x)|^{1/x},$$
where $f(x) = \dfrac x{x^2+1}$ is differentiable and satisfies $f(n) = A_n$.  This allows us to then do this:
\begin{align*}
  \lim_{n\to+\infty} |A_n|^{1/n} &= \lim_{x\to+\infty} |f(x)|^{1/x}\\[0.3cm]
    &= \lim_{x\to+\infty} \left( \frac x{x^2+1} \right)^{1/x} \\[0.3cm]
    &= \lim_{x\to+\infty} \exp\left[\ln\left( \frac x{x^2+1} \right)^{1/x}\right] \\[0.3cm]
    &= \exp\left[\lim_{x\to+\infty} \ln\left( \frac x{x^2+1} \right)^{1/x}\right] \\[0.3cm]
    &= \exp\left[\lim_{x\to+\infty} \frac1x \ln\left( \frac x{x^2+1} \right)\right]
\end{align*}
Now, since direct evaluation of the limit inside the $\exp$ gives us $-\infty/\infty$, this tells us we can use l'Hopital's rule.  So let's do that.
\begin{align*}
  \exp\left[\lim_{x\to+\infty} \frac1x \ln\left( \frac x{x^2+1} \right)\right]  &= \exp\left[\lim_{x\to+\infty} \frac{\ln x - \ln(x^2+1)}x\right]\\[0.3cm]
    &= \exp\left[\lim_{x\to+\infty} \frac{\frac1x - \frac{2x}{x^2+1}}1\right]\\[0.3cm]
    &= \exp(0)\\
    &= 1
\end{align*}
So the root test is inconclusive because $\lim_{n\to+\infty} |A_n|^{1/n} = 1$.
A: Do you insist on using the root test? 
If not, then you could do a limit comparison 
$$
a_n = \frac{n}{n^2+1} \\
b_n = \frac{1}{n}
$$
Here, of course, $\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent. 
If you do insist on the root test, then to find
$$
\lim_{n\to \infty} y
$$
where $y = \sqrt[n]{n^2+1}$ it indeed will work if you let $\ln(y) = \frac{1}{n}\ln(n^2 + 1)$. Then
$$
\lim_{n\to \infty} \frac{\ln(n^2 + 1)}{n} = \lim_{n\to \infty} \frac{2n/(n^2 + 1)}{1} = 0
$$
So
$$
\lim_{n\to \infty} y = 1.
$$
In all the Root test gives you a limit of $1$, so it doesn't work.
