Prove that `.^(n)C_(1) - (1+1/2) .^(n)C_(2) + (1+1/2+1/3) .^(n)C_(3) + "…" ``+ (-1)^(n-1) (1+1/2+1/3 + "…." + 1/n) .^(n)C_(n) = 1/n`

Explanation

Let `S = .^(n)C_(1) - (1+1/2) .^(n)C_(2) + (1+1/2 + 1/3) .^(n)C_(3) + "….."`
`+(-1)^(n-1)(1+1/2+1/3+"…."+1/n) .^(n)C_(n)`.
rth term of the series is
`T(r ) = (-1)^(r-1) . C_(r ) (1+1/2+1/3+"……"+1/r)`, where `C_(r) = .^(n)C_(r)`
Let us consider a series whose general term is
`T_(1)(r ) = (-1)^(r-1). C_(r)(1+x+x^(2)+"...."+x^(r-1))`
`= (-1)^(r-1). C_(r) ((1-x^(r))/(1-x))`
` = ((-1)^(r-1)C_(r))/(1-x) + ((-1)^(r)x^(r)C_(r))/(1-x)`
`rArr underset(r=1)overset(n)sum T_(1)(r ) = (1)/((x-1)) underset(r=1)overset(n)sum (-1)^(r) C_(r ) + (1)/((1+x))underset(r=1)overset(n)sum(-1)^(r). C_(r).x^(r )`
`rArr underset(r=1)overset(n)sumT_(1)(r ) = (1)/((x-1)) (0-x) + (1)/((1-x)) ((1-x)^(n) - 1) = (1-x)^(n-1)`
`= L` (say)
Clearly `S = underset(0 )overset(1)intLdx = underset(0)overset(1)int(1-x)^(n-1) dx = 1/n`