If $a_{n}$ is a non-negative, decreasing sequence, we know from the integral test that if $f(n)=a_{n}$ is an integrable function, then $\sum_{n=1}^{\infty} a_{n}$ and $\int_{1}^{\infty} f(x)dx$ converge/diverge together.
When the theorem is proven, it is shown that:
$\forall k \in \mathbb{N}, a_{k+1} \leq \int_{k}^{k+1} f(x)dx \leq a_{k}$
Which gives us an upper bound of $a_{n}$. Can I use this fact and conclude the following:
$\sum_{n=1}^{\infty} a_{n} \overset{?}{\leq} \int_{1}^{\infty} f(x)dx + a_{1}$
What I'm trying to do is check if $\sum\limits_{n=1}^{\infty} \frac{a}{a^2+n^2} \lt \frac{\pi}{4}+\frac{1}{2}$ when $0\lt a\lt 1$. Using the above gives:
$\int_{1}^{\infty} \frac{a}{a^2+x^2} dx = \frac{1}{a} \int_{1}^{\infty} \frac{1}{1+(\frac{x}{a})^2} dx=\int_{1/a}^{\infty} \frac{1}{1+t^2} dt=$
$=\lim_{t \to \infty} \arctan t - \arctan 1/a \lt \frac{\pi}{2} - 1$
And then
$\sum_{n=1}^{\infty} \frac{a}{a^2+n^2} \lt \frac{\pi}{2} - 1 + \frac{a}{a^2+1} \lt \frac{\pi}{2}$
(For some reason the end result is not what I wanted since $\frac{\pi}{2} \gt \frac{\pi}{4} + 1/2$ but I may have made an error along the way).