Find the supremum of $\left\{ \left\vert \sum^{\infty}_{n=1} \frac{a}{n^{2}+a^{2}} \right\vert: a \in \mathbb{R} \right\}$ I was looking to find the supremum of this set of real numbers
$$\left\{ \left\vert \sum^{\infty}_{n=1} \frac{a}{n^{2}+a^{2}} \right\vert: a \in \mathbb{R} \right\}$$
I was able to show (I hope this is right) that for any $a \in \mathbb{R}$, that 
$$\left\vert \sum^{\infty}_{n=1} \frac{a}{n^{2}+a^{2}} \right\vert< \left\vert \int^{\infty}_{0} \arctan(x/a)dx \right\vert < \frac{\pi}{2}$$
With this I feel that the natural guess for the supremum would be $\pi/2$. But given any $\epsilon>0$, how would one show that $\frac{\pi}{2}-\epsilon$ is not an upperbound?
 A: Put $\displaystyle S(x)=\sum_{n\geq 1}\frac{x}{n^2+x^2}$. Let $N\geq 1$, $m\geq 1$.
We have
$$S(N)\geq \sum_{n=1}^{mN}\frac{N}{n^2+N^2}\geq \sum_{j=0}^{m-1} (\frac{1}{N}\sum_{k=1}^N \frac{1}{1+(j+k/N)^2})=\sum_{j=0}^{m-1} R_j(N)$$
Now $R_j(N)$ is a Riemann sum, and $\displaystyle R_j(N)\to \int_0^1\frac{dt}{1+(j+t)^2}$ as $N\to +\infty$.
If $M$ is an upper bound for $S(x)$, $x\geq 0$, we get 
$$M\geq \sum_{j=0}^{m-1} \int_0^1\frac{dt}{1+(j+t)^2}=\int_0^m \frac{dt}{1+t^2}={\rm Arctan}(m)$$
and it is easy to finish. 
A: I was able to answer the question, a few days after I asked the question. I was a bit lazy typing it up.
The natural guess would be to see that the least upper bound is indeed $\pi/2$. For convenience, the work here deals with when $a \in \mathbb{R}^{>0}$.  To show this is indeed in the affirmative, we show for any $\epsilon>0$ there exists an $a$ where $$\frac{\pi}{2} - \epsilon \leq \sum^{\infty}_{n=1} \frac{a}{n^{2}+a^{2}}$$  this is equivalent to showing 
$$\frac{\pi}{2} -  \sum^{\infty}_{n=1} \frac{a}{n^{2}+a^{2}} \leq \epsilon$$ for some $a$. Our result will rely heavily on the following claim:
For every $\epsilon>0$, there exists an $N$, where for all $m \geq N$, we have that
$$\sum^{\infty}_{m} \frac{a}{a^{2}+n^{2}}+\int^{\infty}_{m} \frac{a}{a^{2}+x^{2}}dx< \epsilon$$
The proof of this will be given later in a more general setting, it is a consequence of the proof of the integral test. 
We now make our argument,
\begin{align*}
\frac{\pi}{2}-\sum^{\infty}_{n=1} \frac{a}{a^{2}+n^{2}} &= \int^{\infty}_{1} \frac{a}{a^{2}+x^{2}}dx- \sum^{\infty}_{n=1} \frac{a}{a^{2}+n^{2}} + \int^{1}_{0} \frac{a}{a^{2}+x^{2}}dx \\ 
&= \Big( \int^{m}_{1} \frac{a}{a^{2}+x^{2}}dx + \int^{\infty}_{m} \frac{a}{a^{2}+x^{2}}dx \Big) \\
&- \Big(\sum^{m}_{n=1} \frac{a}{a^{2}+n^{2}} +\sum^{\infty}_{n=m} \frac{a}{a^{2}+n^{2}} \Big) +  \int^{1}_{0} \frac{a}{a^{2}+x^{2}}dx \\
&= \Big( \int^{m}_{1} \frac{a}{a^{2}+x^{2}}dx -\sum^{m}_{n=1} \frac{a}{a^{2}+n^{2}}  \Big) \\
&- \Big(\int^{\infty}_{m} \frac{a}{a^{2}+x^{2}}dx - \sum^{\infty}_{n=m} \frac{a}{a^{2}+n^{2}} \Big) +  \int^{1}_{0} \frac{a}{a^{2}+x^{2}}dx \\
&= \Big( \int^{m}_{0} \frac{a}{a^{2}+x^{2}}dx -\sum^{m}_{n=1} \frac{a}{a^{2}+n^{2}}  \Big) 
- \Big(\int^{\infty}_{m} \frac{a}{a^{2}+x^{2}}dx - \sum^{\infty}_{n=m} \frac{a}{a^{2}+n^{2}} \Big)  \\
&= \Big( \arctan(m/a) -\sum^{m}_{n=1} \frac{a}{a^{2}+n^{2}}  \Big) 
+\eta   \\ 
\end{align*}
The result then follows from the fact we can choose $a$ large enough so that $\arctan(m/a)$ is small, it is worth nothing that $\Big( \arctan(m/a) -\sum^{m}_{n=1} \frac{a}{a^{2}+n^{2}}  \Big) $ is always positive. The $\eta$ the error associated to the above claim.
Proof of claim: We can actually prove more, given a function $f: [1, \infty) \rightarrow \mathbb{R}$, which is decreasing monotonically yet $f(x) \geq 0$ for all $x \in [1, \infty)$.  The integral test tells us that the sum $$\sum_{n=1}^{\infty} f(n)$$ exists if and only if the improper integral $$\int^{\infty}_{0} f(x)dx$$ exists. We assume further that $f$ has an anti-derivative. As a consequence of the proof of this theorem, we have that the sum $$\sum_{j=1}^{\infty} \alpha_{j}$$ where $$\alpha_{j}=f(j)-\int^{j+1}_{j}f(x)dx$$ converges. 
To see this we note, that the sequence of partial sums is monotonically increasing and bounded.
