Mathematics - Sequence & Series Question with Solution | TestHub

The $n$ -th term of the series is $T_n=\frac{n^2+n-1}{(n+1)!}$. We can rewrite the numerator as $n(n+1)-1$. So, $T_n=\frac{n(n+1)-1}{(n+1)!}=\frac{n(n+1)}{(n+1)!}-\frac{1}{(n+1)!}$. Simplifying further, $\frac{n(n+1)}{(n+1)!}=\frac{n}{n!}=\frac{1}{(n-1)!}$ for $n\ge 1$ (note that $0!=1$). Thus, $T_n=\frac{1}{(n-1)!}-\frac{1}{(n+1)!}$. This is a telescoping series of the form $u_n-u_{n+2}$ where $u_n=\frac{1}{(n-1)!}$. Let's write out the sum $S_N={\sum }_{n=1}^N{T}_n={\sum }_{n=1}^N\left(\frac{1}{(n-1)!}-\frac{1}{(n+1)!}\right)$.$S_N=\left(\frac{1}{0!}-\frac{1}{2!}\right)+\left(\frac{1}{1!}-\frac{1}{3!}\right)+\left(\frac{1}{2!}-\frac{1}{4!}\right)+\cdots +\left(\frac{1}{(N-1)!}-\frac{1}{(N+1)!}\right)$. Most terms cancel out. The sum simplifies to $S_N=\frac{1}{0!}+\frac{1}{1!}-\frac{1}{N!}-\frac{1}{(N+1)!}$. So, $S_N=1+1-\frac{1}{N!}-\frac{1}{(N+1)!}=2-\frac{1}{N!}-\frac{1}{(N+1)!}$. As $N\to \infty$, $\frac{1}{N!}\to 0$ and $\frac{1}{(N+1)!}\to 0$. Therefore, the sum of the infinite series is $S={\lim }_{N\to \infty }{S}_N=2-0-0=2$.