Denominator is too large

Expressing this as an infinite sum, we have:

$\displaystyle S= \sum_{n=1}^{\infty} \frac{n}{\left(n+2\right)!-\left(n+1\right)!-n!}$

$\displaystyle \implies S= \sum_{n=1}^{\infty} \frac{n}{\left(n+2\right) \cdot \left(n+1\right) \cdot n!-\left(n+1\right) \cdot n!-n!}$

$\displaystyle \implies S= \sum_{n=1}^{\infty} \frac{n}{n! \cdot \left[ \left(n+2\right) \cdot \left(n+1\right)-\left(n+1\right)-1\right]}$

$\displaystyle \implies S= \sum_{n=1}^{\infty} \frac{n}{n! \cdot \left[ \left(n+2-1\right) \cdot \left(n+1\right)-1\right]}$

$\displaystyle \implies S= \sum_{n=1}^{\infty} \frac{n}{n! \cdot \left[ \left(n+1\right)^{2}-1\right]} \displaystyle \implies S= \sum_{n=1}^{\infty} \frac{n}{n! \cdot n \cdot \left(n+2\right)}$

$\displaystyle \implies S= \sum_{n=1}^{\infty} \frac{1}{n! \cdot \left(n+2\right)} \implies S= \sum_{n=1}^{\infty} \frac{1}{\frac{\left(n+2\right)!}{\left(n+1\right)}} \implies S= \sum_{n=1}^{\infty} \frac{\left(n+1\right)}{\left(n+2\right)!}$

$\displaystyle \implies S= \sum_{n=2}^{\infty} \frac{n}{\left(n+1\right)!}$

The last sum is pretty easily known to have its value convergent to one, when its first value of the variable is one. As we removed the first term $\displaystyle \frac{1}{\left(1+1\right)!}=\frac{1}{2}$ , this sum converges to $\displaystyle \boxed{S= \frac{1}{2}}$ .

$\sum_{n=3}^{\infty}\frac{n-2}{n!-(n-1)!-(n-2)!}$ $= \sum_{n=3}^{\infty}\frac{n-2}{(n(n-1)-(n-1)-1)(n-2)!}$ $=\sum_{n=3}^{\infty}\frac{1}{n(n-2)!}$ $=\sum_{n=3}^{\infty}\frac{n-1}{n!}$ $=\sum_{n=3}^{\infty}(\frac{1}{(n-1)!}-\frac{1}{n!})$ $=\frac{1}{2}$ , a telescoping sum.

The general term =

n/[(n + 2)! - (n + 1)! - n!] = n/n![(n + 2)(n + 1) - (n + 1) - 1] = 1/n!(n + 2)

= (n + 1)/n!(n + 1)(n +2) = (n +1)/(n + 2)! = [(n + 2) - 1]/(n + 2)!

= 1/(n + 1)! - 1/(n + 2)!

So the given series = 1/2! - 1/3! + 1/3! - 1/4! + 1/4! - 1/5! + 1/5! - ...... = 1/2! = 1/2

I get 1/2 .... Looking at denominator of 2nd term 4!-3!-2!=4(3!)-3!-2!=(3)3!-2!=(3^2-1)2! and taking a look at numerator we see tht it,s 2 and den (3^2-1)2!=3+1)((3-1) (2!) hence 3-1 or 2 is cancelled .and we get 1/(4 2!) and we get the same pattern for every den like for third 1/(5 3!)... Looking at the 2nd refined form multiplying num by 4and div by 4 gives = 3/5! so first term is 2/3! second 3/4! third 4/5! and so on ..Seeing 1 and 2nd term (1st)2/3!=(3-1)/3!=1/2!-1/3! 2nd 3/4!=(4-1)4!=+1/3!-1/4! so we get 1/2!-1/3!+1/3!-1/4! for1st 2 terms =1/2!-1/4! and now breaking 3rd into its constituents the smae way wud cancel -1/4! by adding +1/4! and so on ..We,ll get to infinty the result 1/2!-1/r! where is infintely large hence 1/r! =0 and so answer =1/2

From that series, We can change n/((n+2)!-(n+1)!-(n)! -----> 1/((n+2)(n)!)

sum_(n=1)^infinity 1/((n+2) n!) = 1/2