Senza alcuna fine.
Inviato: 11 ott 2019, 14:58
Sia [math].
Dimostrare che:
[math]
Dimostrare che:
[math]
Testo nascosto:
[math]\displaystyle \sum_{k=1}^{\infty} \frac{1}{\prod_{j=0}^m (k+j)}=\lim_{n\to +\infty}\sum_{k=1}^{n} \frac{1}{\prod_{j=0}^m (k+j)}
[math]=\displaystyle \lim_{n\to +\infty}\sum_{k=1}^{n}{\frac{(k-1)!}{(k+m)!}}=\lim_{n\to +\infty}[\frac{0!}{(m+1)!}+\frac{1!}{(m+2)!}+ \dots +\frac{(n-1)!}{(n+m)!}]
[math]=\displaystyle \lim_{n\to +\infty}[(\frac{0!}{m \cdot m!}-\frac{1!}{m \cdot (m+1)!})+(\frac{1!}{m \cdot (m+1)!}-\frac{2!}{m \cdot (m+2)!})+ \dots +(\frac{(n-1)!}{m \cdot (m+n-1)!}-\frac{n!}{m \cdot (m+n)!})
[math]=\displaystyle \lim_{n\to +\infty}(\frac{0!}{m \cdot m!}-\frac{n!}{m \cdot (m+n)!})=\frac{1}{m \cdot m!}-\lim_{n\to +\infty} \frac{1}{m(n+1) \dots (n+m)}
[math]=\displaystyle \frac{1}{m \cdot m!}-0=\frac{1}{m \cdot m!}.
Si è fatto uso dell'identità
[math]\displaystyle \frac{i!}{(m+i+1)!}=\frac{i!}{m \cdot (m+i)!}-\frac{(i+1)!}{m \cdot (m+i+1)!}.
[math]=\displaystyle \lim_{n\to +\infty}\sum_{k=1}^{n}{\frac{(k-1)!}{(k+m)!}}=\lim_{n\to +\infty}[\frac{0!}{(m+1)!}+\frac{1!}{(m+2)!}+ \dots +\frac{(n-1)!}{(n+m)!}]
[math]=\displaystyle \lim_{n\to +\infty}[(\frac{0!}{m \cdot m!}-\frac{1!}{m \cdot (m+1)!})+(\frac{1!}{m \cdot (m+1)!}-\frac{2!}{m \cdot (m+2)!})+ \dots +(\frac{(n-1)!}{m \cdot (m+n-1)!}-\frac{n!}{m \cdot (m+n)!})
[math]=\displaystyle \lim_{n\to +\infty}(\frac{0!}{m \cdot m!}-\frac{n!}{m \cdot (m+n)!})=\frac{1}{m \cdot m!}-\lim_{n\to +\infty} \frac{1}{m(n+1) \dots (n+m)}
[math]=\displaystyle \frac{1}{m \cdot m!}-0=\frac{1}{m \cdot m!}.
Si è fatto uso dell'identità
[math]\displaystyle \frac{i!}{(m+i+1)!}=\frac{i!}{m \cdot (m+i)!}-\frac{(i+1)!}{m \cdot (m+i+1)!}.