Ars Mathematica: Solución 1 al problema 80

D_{n} = \frac{r}{n (n + 1)}

d_{n} = \frac{D_{n}}{2}

\begin{matrix} (r + d_{1})^{2} & = & (r - d_{1})^{2} + r^{2} \\ r^{2} + 2 r d_{1} + d_{1}^{2} & = & r^{2} - 2 r d_{1} + d_{1}^{2} + r^{2} \\ 4 r d_{1} & = & r^{2} \\ d_{1} & = & \frac{r}{4} \\ D_{1} & = & \frac{r}{2} \end{matrix}

D_{n - 1} = \frac{r}{n (n + 1)}

\begin{matrix} D_{1} + D_{2} + \dots + D_{n - 1} & = & \frac{r}{1 \cdot 2} + \frac{r}{2 \cdot 3} + \dots + \frac{r}{(n - 1) \cdot n} = \end{matrix}

= r (\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} + \dots + \frac{1}{n - 1} - \frac{1}{n}) = r (1 - \frac{1}{n}) = r \frac{n - 1}{n}

r - D_{1} - D_{2} - \dots - D_{n - 1} - \frac{D_{n}}{2} = r - r \frac{n - 1}{n} - \frac{D_{n}}{2} = \frac{r}{n} - \frac{D_{n}}{2}

\begin{matrix} {(r + \frac{D_{n}}{2})}^{2} & = & r^{2} + {(\frac{r}{n} - \frac{D_{n}}{2})}^{2} \\ r^{2} + r D_{n} + D_{n}^{2} & = & r^{2} + \frac{r^{2}}{n^{2}} - \frac{r D_{n}}{n} + \frac{D_{n}^{2}}{4} \\ r D_{n} (1 + \frac{1}{n}) & = & \frac{r^{2}}{n^{2}} \\ D_{n} (\frac{n + 1}{n}) & = & \frac{r}{n^{2}} \\ D_{n} & = & \frac{r n}{n^{2} (n + 1)} = \frac{r}{n (n + 1)} \end{matrix}

r = 1

D_{1} + D_{2} + \dots + D_{n} = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n (n + 1)}

n

lim_{n \to \infty} (\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n \cdot (n + 1)}) = lim_{n \to \infty} (D_{1} + D_{2} + \dots + D_{n}) = 1

Ars Mathematica