Answer: (i) \(\dfrac{n(\mathrm e-1)+\mathrm e}{n(n+1)\mathrm e^{n+1}}=\dfrac{1}{n\mathrm e^n}-\dfrac{1}{(n+1)\mathrm e^{n+1}}\).

(ii) \(S_N=\dfrac{1}{\mathrm e}-\dfrac{1}{(N+1)\mathrm e^{N+1}}\), so \(S=\dfrac{1}{\mathrm e}\).

(iii) The least value is \(N=6\).

For the identity, put the right-hand side over the common denominator \(n(n+1)\mathrm e^{n+1}\):

\(\dfrac{1}{n\mathrm e^n}-\dfrac{1}{(n+1)\mathrm e^{n+1}}=\dfrac{(n+1)\mathrm e-n}{n(n+1)\mathrm e^{n+1}}=\dfrac{n(\mathrm e-1)+\mathrm e}{n(n+1)\mathrm e^{n+1}}.\)

Therefore

\(S_N=\sum_{n=1}^{N}\left(\dfrac{1}{n\mathrm e^n}-\dfrac{1}{(n+1)\mathrm e^{n+1}}\right).\)

This is a telescoping sum:

\(S_N=\left(\dfrac1{\mathrm e}-\dfrac1{2\mathrm e^2}\right)+\left(\dfrac1{2\mathrm e^2}-\dfrac1{3\mathrm e^3}\right)+\cdots+\left(\dfrac1{N\mathrm e^N}-\dfrac1{(N+1)\mathrm e^{N+1}}\right).\)

All intermediate terms cancel, so

\(S_N=\dfrac1{\mathrm e}-\dfrac1{(N+1)\mathrm e^{N+1}}.\)

Hence

\(S=\lim_{N\to\infty}S_N=\dfrac1{\mathrm e}.\)

Then

\(S-S_N=\dfrac1{(N+1)\mathrm e^{N+1}}\),

so

\((N+1)(S-S_N)=\dfrac1{\mathrm e^{N+1}}.\)

We need \(\mathrm e^{-(N+1)}\lt 10^{-3}\). Since \(\ln 1000\approx 6.91\), we need \(N+1>6.91\), so the least integer value is

\(N=6\).