Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $\pi/2$.

share|cite|improve this answer
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.

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.