= Solution
The special <binomial theorem>
$$
(1+t)^n=\sum_{k=0}^n\binom nk t^k
$$
follows by <mathematical induction>. Multiplication of the formula for $n$ by $1+t$ and collection of the coefficient of $t^k$ uses Pascal's identity from part (a).
Termwise integration from $0$ to $1$ gives
$$
\boxed{
\sum_{k=0}^n\frac1{k+1}\binom nk
=\int_0^1(1+t)^n\,dt
=\frac{2^{n+1}-1}{n+1}}.
$$
Replacing $t$ by $-t$ gives
$$
\boxed{
\sum_{k=0}^n\frac{(-1)^k}{k+1}\binom nk
=\int_0^1(1-t)^n\,dt
=\frac1{n+1}}.
$$
Finally,
$$
\sum_{k=1}^n\frac{(-1)^{k+1}}k\binom nk
=\int_0^1\frac{1-(1-x)^n}{x}\,dx.
$$
Writing $y=1-x$ and using the finite <geometric series>,
$$
\frac{1-y^n}{1-y}=1+y+\cdots+y^{n-1},
$$
turns this into
$$
\boxed{\sum_{j=0}^{n-1}\int_0^1y^j\,dy
=1+\frac12+\cdots+\frac1n}.
$$
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