Bounding a series from above using the integral test 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).

Given $\forall k \in \mathbb{N}, a_{k+1} \leq \int_{k}^{k+1} f(x)dx \leq a_{k}$, 
$$\sum_{k=1}^{\infty}a_{k+1} \leq \sum_{k=1}^{\infty}\int_{k}^{k+1} f(x)dx \leq \sum_{k=1}^{\infty}a_{k}$$
so
$$\sum_{k=2}^{\infty}a_{k} \leq \int_{1}^{\infty} f(x)dx \leq \sum_{k=1}^{\infty}a_{k}$$
and, using the left part,
$$\sum_{k=1}^{\infty}a_{k}=a_1+\sum_{k=2}^{\infty}a_{k} \leq a_1+\int_{1}^{\infty} f(x)dx.$$
Now, to $\int_{1}^{\infty} \frac{a}{a^2+x^2} dx=\frac{\pi}{2}-\arctan\frac{1}{a}$.  As in Jack Schmidt's comment, $\arctan\frac{1}{a}>\frac{\pi}{4}$ when $0<a<1$, so $\frac{\pi}{2}-\arctan\frac{1}{a}<\frac{\pi}{4}$.  So,
$$\sum_{k=1}^{\infty}a_{k}\leq a_1+\int_{1}^{\infty} f(x)dx<\frac{a}{a^2+1}+\frac{\pi}{4}<\frac{1}{2}+\frac{\pi}{4}.$$
