Answer
We prove:
$\mathop \sum \limits_{n = 0}^\infty n{{\rm{e}}^{ - nx}} = \dfrac{{{{\rm{e}}^{ - x}}}}{{{{\left( {1 - {{\rm{e}}^{ - x}}} \right)}^2}}}$
Work Step by Step
We notice that $\dfrac{d}{{dx}}\left( { - \dfrac{1}{{1 - {{\rm{e}}^{ - x}}}}} \right) = \dfrac{{{{\rm{e}}^{ - x}}}}{{{{\left( {1 - {{\rm{e}}^{ - x}}} \right)}^2}}}$.
Consider the expansion of the function $\dfrac{1}{{1 - {{\rm{e}}^{ - x}}}}$. Taking the derivative of this expansion, we shall obtain the series representing $\dfrac{{{{\rm{e}}^{ - x}}}}{{{{\left( {1 - {{\rm{e}}^{ - x}}} \right)}^2}}}$.
Substitute ${{\rm{e}}^{ - x}}$ for $x$ in Eq. (2) of Section 11.6 gives
$\dfrac{1}{{1 - {{\rm{e}}^{ - x}}}} = \mathop \sum \limits_{n = 0}^\infty {\left( {{{\rm{e}}^{ - x}}} \right)^n} = \mathop \sum \limits_{n = 0}^\infty {{\rm{e}}^{ - nx}}$, ${\ \ \ }$ converges for $\left| {{{\rm{e}}^{ - x}}} \right| \lt 1$
Taking the derivative on the series with respect to $x$ gives
$\dfrac{d}{{dx}}\left( {\dfrac{1}{{1 - {{\rm{e}}^{ - x}}}}} \right) = - \mathop \sum \limits_{n = 0}^\infty n{{\rm{e}}^{ - nx}}$
Hence,
$\dfrac{d}{{dx}}\left( { - \dfrac{1}{{1 - {{\rm{e}}^{ - x}}}}} \right) = \mathop \sum \limits_{n = 0}^\infty n{{\rm{e}}^{ - nx}}$
Since $\dfrac{d}{{dx}}\left( { - \dfrac{1}{{1 - {{\rm{e}}^{ - x}}}}} \right) = \dfrac{{{{\rm{e}}^{ - x}}}}{{{{\left( {1 - {{\rm{e}}^{ - x}}} \right)}^2}}}$, therefore $\mathop \sum \limits_{n = 0}^\infty n{{\rm{e}}^{ - nx}} = \dfrac{{{{\rm{e}}^{ - x}}}}{{{{\left( {1 - {{\rm{e}}^{ - x}}} \right)}^2}}}$.
