$1(1!) + 2(2!) + 3(3!) + \dots + n(n!) = \dots$

The sum of the series $\frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots$ up to $n$ terms is equal to:

What is the sum of the first $n$ terms of the series whose $k$-th term is $k! \times k$?

If $\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots \left(1+\frac{2n+1}{n^2}\right) = 121$,then $n =$

If the sum of the first $10$ terms of the series $\frac{4(1)}{1+4(1)^4}+\frac{4(2)}{1+4(2)^4}+\frac{4(3)}{1+4(3)^4}+\ldots$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $m+n$ is equal to . . . . . . .

Find the sum of the series: $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots + \frac{1}{n(n + 1)}$