# About this limit, $\lim_{t\to0^+} \sum_{n=1}^\infty \frac{\sqrt t}{1 + tn^2}$

This recent question, Evaluating a limit, 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?

-
Questions should be self-contained... – Listing Aug 17 '11 at 18:27
Yes, the question should be rewritten. But the function is not continuous at $0$, because its value at $0$ is $0$, and the function tends to $\frac{\pi}{2}$ as $t$ tends to 0 from above. For a function $f$ to be continuous at $a,$ we need $\lim_{t \to a} f(t) = f(a).$ – Geoff Robinson Aug 17 '11 at 18:34

The function $\sqrt{t}$ is not defined for negative $t$. So it makes no sense to look at the behaviour as $t$ approaches $0$ through negative values.

The sum, as a function $F(t)$, can be expressed, for $t>0$, in terms of the hyperbolic cotangent of $\pi/\sqrt{t}$. For $t=0$, each term is $0$, so $F(0)=0$.

It was shown by @Shai Covo and @anon that $\lim_{t\to 0+}F(t)=\pi/2$. So $F(t)$ is not continuous (from the right) at $t=0$.

However, if we define the function $G(t)$ by $G(t)=F(t)$ if $t>0$, and $G(0)=\pi/2$, then the function $G$ is continuous (from the right) at $t=0$. This is because $\lim_{t\to 0+}F(t)=\pi/2$.

Of course $G(t)$ cannot be fully continuous at $0$, since we have not even defined it for negative $t$. If we cared to, we could define $H(t)$ by $H(t)=G(t)$ if $t\ge 0$, and $H(t)=G(|t|)$ for $t<0$. Then $H(t)$ would be continuous for all $t$. This is not all that unreasonable. Instead of summing $\sqrt{t}/(1+n^2 t)$, we would be summing $\sqrt{|t|}/(1+n^2 |t|)$.

The discontinuity (from the right) of $F(t)$ at $t=0$ is in a sense not a mathematically significant one. The technical term is that it is a removable discontinuity. The value of $F(0)$ is the "wrong one" for continuity from the right, but that can be easily changed by replacing that value by the "correct one," which should be $\pi/2$.

-
For negative $t$ the $\sqrt{t}$ can be understood as a complex number; more serious is the $1+ t n^2$ is the denominator, which makes the $n$th term undefined at $t = -1/n^2$. These discontinuities are not removable. The limit as $t \to 0-$ does not exist because there is no interval $(-\epsilon,0)$ on which your function is defined. – Robert Israel Aug 17 '11 at 23:29

The "significance" is that the sum is actually a Riemann sum that approximates the integral $\int_0^\infty \frac{dx}{1+x^2}$ with interval lengths $\sqrt{t}$. Consequently as $t\to 0+$, the sum approaches the integral.

It is in fact discontinuous from the right. If $t$ is actually equal to $0$, then the sum is exactly $0$, but as $t$ approaches $0$, the sum approaches $\pi/2$, not $0$. That is a discontinuity. But that does not explain why the limit is one-sided. What one would do with $\sqrt{t}$ if $t$ were negative is not altogether clear, and at any rate a negative number cannot be the length of the intervals in a Riemann sum.

-
Thanks. This is what I was looking for. Problems like this do occur in Cal I, II exercises disguised for Riemann integrals. – clkirksey Aug 18 '11 at 15:42
The Riemann sum, like the proper Riemann integral, require a finite domain of integration. Usually, the infinite domains work out, but it's nice to make sure. – robjohn Jan 16 '13 at 4:06

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 $$\lim_{k\to\infty}\sum_{n=1}^\infty\frac{1/k}{1+{(n/k)^2}} =\int_0^\infty\frac{\mathrm{d}x}{1+x^2} =\frac\pi2$$

-