Rules for Integral Test for Convergence/Divergence Suppose we're considering the series:
$$\sum_{n=2}^\infty\frac{n+\sin n}{1+n^2}$$
For some reason, it seems I can't use the integral test for this problem despite the fact that it fulfills the criteria of being positive, continuous and decreasing. What exactly would the reason be?
 A: $\displaystyle\sum\dfrac{n}{1+n^{2}}\geq\sum\dfrac{n}{n^{2}+n^{2}}=\dfrac{1}{2}\sum\dfrac{1}{n}=\infty$.
$\displaystyle\sum\dfrac{|\sin n|}{1+n^{2}}\leq\sum\dfrac{1}{n^{2}}<\infty$.
If it were convergent, then $\displaystyle\sum\dfrac{n}{1+n^{2}}=\sum\left(\dfrac{n}{1+n^{2}}+\dfrac{\sin n}{1+n^{2}}\right)-\sum\dfrac{\sin n}{1+n^{2}}$ would be convergent, a contradiction.
A: The summand can be bounded:
$n+\sin(n) \geq n-1 \geq \frac{n}{2} ~\forall n \geq 2$
$1+n^2 < 2n^2 ~\forall n \geq 1$
Therefore, the sum is lower-bounded by $\sum_n \frac{1}{4n}$ which is $\frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n}$, which diverges, because the harmonic series does. 
A: You "shouldn't" use the integral test because the integral $\int \frac{x + \sin x}{x^2+1}\, dx$ is difficult.
You "can't" use the integral test because the integrand, 
$\frac{x + \sin x}{x^2+1}$, is not monotone as $x\to \infty$.
The derivative of $\frac{x + \sin x}{x^2+1}$ is $\frac{1+\cos x -2x\sin x + x^2(\cos x -1)}{(1+x^2)^2}$ which has the same sign as the numerator
$$
1+\cos x -2x\sin x + x^2(\cos x -1)
$$ When $\cos x$ is $1$ this expresion is $2$ and the integrand is increasing.  So the integral test does not apply.  Fortunately the limit comparison test easily gives confirmation of divergence.

The limit comparison test seems appropriate: $\frac{n+\sin n}{1+n^2}\sim \frac{1}{n}$.
$$
\lim_{n\to\infty} \frac{\frac{n+\sin n}{1+n^2}}{\frac{1}{n}}
=\lim_{n\to\infty}\frac{n^2+n\sin n}{1+n^2}
=\lim_{n\to\infty}\frac{1+\sin n/n}{1/n^2+1}
=1
$$
and since $\sum\frac{1}{n}$ diverges, so does $\sum\frac{n+\sin n}{1+n^2}$.
