We can construct a Dirac sequence $\{\delta_n\}_{n\in \mathbb{N}_+}$ by choosing $$ \delta_n(x) = \frac{n}{\pi(1+n^2x^2)}. $$ To show that conditions 1 and 3 in the definition of a Dirac sequence are satisfied is trivial and condition 2 is also fulfilled since $$ \int_{-\infty}^{\infty}\delta_n (x)dx = \frac{1}{\pi}\cdot \int_{-\infty}^{\infty} \frac{n}{1+n^2x^2} dx = \left[\begin{tabular}{ll}$y=nx$ \\ $dy=n\cdot dx$ \end{tabular}\right]= \frac{1}{\pi}\cdot\int_{-\infty}^{\infty} \frac{1}{1+y^2} dy = \frac{1}{\pi}\cdot \arctan{y}\Big\lvert_{y=-\infty}^{\infty} =\frac{1}{\pi}\cdot\pi =1 $$ for all $n\in\mathbb{N}_+$ , hence $\{\delta_n\}_{n\in\mathbb{N}_+}$ is a Dirac sequence.

To prove that it actually converges in $\mathcal{D}'(\mathbb{R})$ (the space of all distributions on $\mathcal{D}(\mathbb{R})$ ) to the Dirac delta distribution $\delta$ , we must show that $$ \lim_{n\to \infty} \int_{\mathbb{R}} \delta_n(x)\varphi(x)dx = \varphi(0) $$ for any test function $\varphi\in\mathcal{D}(\mathbb{R})$ (a topological vector space of smooth functions with compact support). Let us take an arbitrary test function $\varphi\in\mathcal{D}(\mathbb{R})$ and assume that the closed and compact set $\supp(\varphi)$ is contained in some open interval $(a,b)\subset \mathbb{R}$ ($a<0$ and $b>0$ ). Using the triangle inequality and the fact that $\int_{\mathbb{R}}\delta_n(x)dx=1$ for all $n\in\mathbb{N}_+$ we can write $$ \left| \int_{-\infty}^{\infty}\delta_n(x)\varphi(x)dx -\varphi(0) \right| = \left|\int_{-\infty}^{\infty}\delta_n(x)(\varphi(x)-\varphi(0))dx \right| \leq $$ $$ \leq \underbrace{\varphi(0) \int_{-\infty}^{a}\left|\delta_n(x)\right|dx}_{I_1} + \underbrace{\int_{a}^{b}\left|\delta_n(x)(\varphi(x)-\varphi(0))\right|dx}_{I_2} + \underbrace{\varphi(0) \int_{b}^{\infty}\left|\delta_n(x)\right|dx}_{I_3} $$ It is easy to see that $\lim_{n\to \infty} \delta_n(x)=0$ , $\forall x\in (-\infty,a] \cup [b,\infty)$ and therefore $\lim_{n\to \infty} I_1 = 0$ and $\lim_{n\to \infty} I_3 = 0$ . Finally we want to estimate $I_2$ when $n\to\infty$ . $$ I_2 = \int_{a}^{b}\left|\delta_n(x)\right| \underbrace{\left|(\varphi(x)-\varphi(0))\right|}_{\leq |x|\cdot \sup{|\varphi'(x)|}}dx \leq \sup{|\varphi'(x)|} \cdot \int_{a}^{b}\left|\delta_n(x)x \right| dx = $$ $$ = \sup{|\varphi'(x)|} \cdot \frac{1}{\pi}\int_{a}^{b}\left|\frac{nx}{1+(nx)^2}\right| dx = \sup{|\varphi'(x)|} \cdot \frac{1}{\pi}\left( -\int_{a}^{0}\frac{nx}{1+(nx)^2} dx +\int_{0}^{b}\frac{nx}{1+(nx)^2} dx \right)= $$ $$ = \sup{|\varphi'(x)|} \cdot \frac{1}{\pi}\left( - \left(\frac{1}{2n}\cdot \text{ln}|1+(nx)^2| \Big\lvert_{x=a}^{0}\right) +\left( \frac{1}{2n}\cdot \text{ln}|1+(nx)^2| \Big\lvert_{x=0}^{b} \right) \right)= $$ $$ = \sup{|\varphi'(x)|} \cdot \frac{1}{\pi}\left( \frac{1}{2n}\cdot \text{ln}|1+(na)^2| + \frac{1}{2n}\cdot \text{ln}|1+(nb)^2| \right) $$ We now conclude that $\lim_{n\to \infty} I_2 = 0$ . This means that $\lim_{n\to\infty}I_1+I_2+I_3=0$ which shows that $\{ \delta_n \}_{n\in\mathbb{N}_+}$ converges to the Dirac delta distribution $\delta$ .