Answer: We prove by induction that for all positive integers \(N\), \(S_N=1-\frac{1}{(N+1)!}\).

Let \(H_N\) be the statement \(S_N=1-\frac{1}{(N+1)!}\), where \(S_N=\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{N}{(N+1)!}\).

Base case: For \(N=1\),

\(S_1=\frac{1}{2!}=\frac12\).

Also, \(1-\frac{1}{(1+1)!}=1-\frac{1}{2!}=1-\frac12=\frac12\).

So \(H_1\) is true.

Inductive step: Assume that for some \(k\ge 1\),

\(S_k=1-\frac{1}{(k+1)!}\).

We must show that \(S_{k+1}=1-\frac{1}{(k+2)!}\).

Now

\(S_{k+1}=S_k+\frac{k+1}{(k+2)!}\).

Using the inductive hypothesis,

\(S_{k+1}=1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!}\).

Since \(\frac{1}{(k+1)!}=\frac{k+2}{(k+2)!}\), this becomes

\(S_{k+1}=1-\frac{k+2}{(k+2)!}+\frac{k+1}{(k+2)!}\).

So

\(S_{k+1}=1-\frac{1}{(k+2)!}\).

Therefore \(H_k \Rightarrow H_{k+1}\).

By mathematical induction, \(S_N=1-\frac{1}{(N+1)!}\) for all positive integers \(N\).