TITLE: About this limit, $\lim_{t\to0^+} \sum_{n=1}^\infty \frac{\sqrt t}{1 + tn^2}$
QUESTION [6 upvotes]: This recent question, Evaluating a limit, $\lim\limits_{t\rightarrow 0^{+}} {\sum\limits_{n=1} ^{\infty} \frac{\sqrt{t}}{1+tn^2}}$, asked for the value of
$$\lim_{t\to0^+} \sum_{n=1}^\infty \frac{\sqrt t}{1 + tn^2}$$
So that I could better understand the answer can someone explain if this function of $t$ is discontinuous at $t=0$ and that is why the right-sided limit has to be taken? Does this function have any significance?

REPLY [2 votes]: To make the notation simpler, substitute $t\mapsto t^2$:
$$
\lim_{t\to0^+}\sum_{n=1}^\infty\frac{\sqrt t}{1+tn^2}
=\lim_{t\to0^+}\sum_{n=1}^\infty\frac{t}{1+{(tn)^2}}
$$
First note that
$$
\begin{align}
\sum_{n=N/t}^\infty\frac{t}{1+{(tn)^2}}
&\le\frac1t\sum_{n=N/t+1}^\infty\frac1{n^2}\\
&\le\frac1t\sum_{n=N/t+1}^\infty\frac1{n(n-1)}\\
&=\frac1N
\end{align}
$$
Since Riemann integrals are only over finite intervals, we need to restrict the sum to get the Riemann Sum (where $tn=x$ and $t=\mathrm{d}x$)
$$
\begin{align}
\lim_{t\to0^+}\sum_{n=1}^{N/t}\frac{t}{1+{(tn)^2}}
&=\int_0^N\frac{\mathrm{d}x}{1+x^2}\\
&=\arctan(N)
\end{align}
$$
Therefore, letting $N\to\infty$,
$$
\begin{align}
\lim_{t\to0^+}\sum_{n=1}^\infty\frac{t}{1+{(tn)^2}}
&=\arctan(N)+O\left(\frac1N\right)\\
&=\frac\pi2
\end{align}
$$

Alternate approach
Set $t=1/k$, then
$$
\lim_{t\to0^+}\sum_{n=1}^\infty\frac{\sqrt t}{1+tn^2}
=\lim_{t\to0^+}\sum_{n=1}^\infty\frac{t}{1+{(tn)^2}}
=\lim_{k\to\infty}\sum_{n=1}^\infty\frac{1/k}{1+{(n/k)^2}}
$$
Since $\frac{1/k}{1+{(n/k)^2}}$ is monotonically decreasing (in $n$), the Integral Test says
$$
\int_0^\infty\frac{\mathrm{d}x}{1+x^2}
\le\sum_{n=1}^\infty\frac{1/k}{1+{(n/k)^2}}
\le\int_{1/k}^\infty\frac{\mathrm{d}x}{1+x^2}
$$
By the Squeeze Theorem, we get
$$
\lim_{k\to\infty}\sum_{n=1}^\infty\frac{1/k}{1+{(n/k)^2}}
=\int_0^\infty\frac{\mathrm{d}x}{1+x^2}
=\frac\pi2
$$