Rules for Integral Test for Convergence/Divergence

Question

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?

Answer 1

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 $\sum\frac{n+\sin n}{1+n^2}$.

Answer 2

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.

Answer 3

$\displaystyle\sum\dfrac{n}{1+n^{2}}\geq\sum\dfrac{n}{n^{2}+n^{2}}=\dfrac{1}{2}\sum\dfrac{1}{n}=\infty$.

$\displaystyle\sum\dfrac{|\sin n|}{1+n^{2}}\leq\sum\dfrac{1}{n^{2}}<\infty$.

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.