Be a reverse engineer

The given summation can be rewritten as -

$S = \sum_{k=0}^n {n \choose k}\frac{(-1)^k}{k+1}$

Observe that -

$\frac{1}{n+1} = \int_0^1 x^n \mathrm{d}x$

Thus, we can easily evaluate the sum by converting it to integral as follows -

$S = \sum_{k=0}^n {n \choose k}(-1)^k\int_0^1 x^k \mathrm{d}x$

Interchanging the integral and sum gives us -

$S = \int_0^1 \sum_{k=0}^n {n \choose k} (-x)^k \mathrm{d}x$

(We can interchange the integral with sum because the variables are independent)

The above written sum is a binomial sum, according to which -

$(1-x)^n = \sum_{k=0}^n {n \choose k} (-x)^k$

Thus, we get

$S = \int_0^1 (1-x)^n \mathrm{d}x = -\left(\frac{(1-x)^{n+1}}{n+1}\right|_0^1 = \frac{1}{n+1}$

But we have $S = 0.00125 = \frac{1}{n+1} \Rightarrow n = \frac{1}{0.00125}-1 = 800-1$

And hence, $\boxed{n=799}$

L.H.S = $\frac{n!}{n!}-\frac{1}{2}\frac{n!}{(n-1)!}+\frac{1}{3}\frac{n!}{(n-2)!2!}+...+\frac{(-1)^{n}}{n+1}\frac{n!}{n!}$

$=\frac{1}{n+1}\frac{(n+1)!}{n!1!}-\frac{1}{n+1}\frac{(n+1)!}{(n-1)!2!}+\frac{1}{n+1}\frac{(n+1)!}{(n-2)!3!}-....+\frac{(-1))^{n}}{n+1}\frac{(n+1)!}{(n+1)!0!}$

$=\frac{1}{n+1}(\binom{n+1}{1}-\binom{n+1}{2}+\binom{n+1}{3}-...+(-1)^{n}\binom{n+1}{n+1}).....(1)$

Now we have

$(a-b)^{n+1}=a^{n+1}+\sum_{k=1}^{n+1}\binom{n+1}{k}(-1)^{k}a^{n-k+1}b^{k}$

Letting $a=b=1$ in above equation we get $\binom{n+1}{1}-\binom{n+1}{2}+\binom{n+1}{3}-...+(-1)^{n}\binom{n+1}{n+1}=1...(2)$

Substituting $(2)$ in $(1)$ gives answer as $\frac{1}{n+1}$

Therefore $\frac{1}{n+1}=0.00125$

$\Rightarrow n=799$

Whether we see it determined.

Sum (n) = 1/ (n + 1)

0.00125 = 1/ (799 + 1)

n = 799