Series Demystified 9

Let us demystify the expression which is to be summated. $\ddot \smile$

$\dfrac{1+n+n^2}{(n^2+n)(n+1)!} = \dfrac{(1+n)^2 - n}{(n^2+n)(n+1)!}$

$\Rightarrow \dfrac{(1+n)^2}{n(n+1)(n+1)!} - \dfrac{n}{n(n+1)(n+1)!}$

$\Rightarrow \dfrac{(1+n)}{n(n+1)n!} - \dfrac{1}{(n+1)(n+1)!}$

$\Rightarrow \dfrac{1}{n\times n!} - \dfrac{1}{(n+1) \times (n+1)!}$

Now we have a telescopic series

$S = \displaystyle \sum_{n=1}^{48} \dfrac{1}{n\times n!} - \dfrac{1}{(n+1) \times (n+1)!}$

$S = \dfrac{1}{1\times 1!} - \dfrac{1}{2\times 2!} + \dfrac{1}{2\times 2!} - \dfrac{1}{3 \times 3!} + \ldots + \dfrac{1}{48\times 48!} - \dfrac{1}{49\times 49!}$

$S = 1 - \dfrac{1}{49\times 49!}$

$\lceil S \rceil = 1 \Rightarrow \lceil S \rceil - S = 1 - \left(1 - \dfrac{1}{49 \times 49!} \right) = \dfrac{1}{49 \times 49!}$

Hence the expression we want, which is

$\dfrac{1}{\sqrt{49!\left(\lceil S \rceil - S \right)}} = \dfrac{1}{\sqrt{\frac{1}{49}}} = \dfrac{1}{\frac{1}{7}} = \boxed{7}$

Observe that the denominator is $(n)(n+1)(n+1)!$ .

Therefore, we propose that:

$\frac{1+n+n^2}{(n^2+n)(n+1)!} \equiv \frac{An+B}{n(n+1)!} + \frac{Cn+D}{(n+1)(n+1)!}$

Multiplying both sides by $n(n+1)(n+1)!$ :

$1+n+n^2 \equiv B + (A+B+D)n + (A+C)n^2$

Then, comparing like terms give us $B=1$ :

$\frac{1+n+n^2}{(n^2+n)(n+1)!} \equiv \frac{An+1}{n(n+1)!} + \frac{Cn+D}{(n+1)(n+1)!}$

In order to create a telescoping series, The first denominator should be $n\times n!$ while the second denominator is unchanged. Therefore, let $A=1$ .

Comparing like terms give us $C=0$ and $D=-1$ :

$\frac{1+n+n^2}{(n^2+n)(n+1)!} \equiv \frac{n+1}{n(n+1)!} - \frac{1}{(n+1)(n+1)!} \equiv \frac{1}{n\times n!} - \frac{1}{(n+1)(n+1)!}$

Thus a telescoping series has been created.