3. The sequence $\left(a_{n}\right)$ is defined by $a_{1}=\frac{1}{2}, a_{n}=\frac{a_{n-1}}{2 n a_{n-1}+1}$ for $n>1$. Determine the sum $a_{1}+\ldots+a_{k}$, for any natural number $k$.
Solution. First method. Since $a_{1}=\frac{1}{2}>0$, it follows that $a_{n}>0$ for every $n \in \mathbb{N}$. Furthermore, for $n>1$ we have $a_{n}=\frac{a_{n-1}}{2 n a_{n-1}+1}=\frac{1}{2 n+\frac{1}{a_{n-1}}}$. Therefore,

$$
\begin{aligned}
\frac{1}{a_{n}}= & 2 n+\frac{1}{a_{n-1}}=2 n+2(n-1)+\frac{1}{a_{n-2}}=\ldots=2 n+2(n-1)+\ldots+2 \cdot 2+\frac{1}{a_{1}} \\
= & 2 n+2(n-1)+\ldots+2 \cdot 2+2 \cdot 1=2 \frac{n(n+1)}{2}=n(n+1)
\end{aligned}
$$

Accordingly, $a_{n}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$, for every $n \in \mathbb{N}$. Thus, for any $k \in \mathbb{N}$, we have $a_{1}+\ldots+a_{k}=1-\frac{1}{k+1}=\frac{k}{k+1}$

Second method. Using the principle of mathematical induction, we will prove that $a_{k}=\frac{1}{k(k+1)}$, for every $k \in \mathbb{N}$. For $k=1$, it holds that $a_{1}=\frac{1}{1 \cdot 2}$. Assume that the statement holds for the natural number $n-1$. Then for $n$ we have

$$
a_{n}=\frac{a_{n-1}}{2 n a_{n-1}+1}=\frac{\frac{1}{n(n-1)}}{2 n \frac{1}{n(n-1)}+1}=\frac{1}{2 n+n^{2}-n}=\frac{1}{n(n+1)}
$$

so $a_{k}=\frac{1}{k(k+1)}$, for every $k \in \mathbb{N}$.

Finally, $a_{1}+\ldots+a_{k}=1-\frac{1}{k+1}=\frac{k}{k+1}$