Recall that, in general,

\(\displaystyle{1}+{x}+{x}^{{2}}+\ldots={\frac{{{1}}}{{{1}-{x}}}},\ {\left|{x}\right|}{ < }{1}\)

Moreover, power series can be differentiated term by term. So, differentiating both sides of the equation above we get

\(\displaystyle{1}+{2}{x}+{3}{x}^{{2}}+\ldots={\frac{{{1}}}{{{\left({1}-{x}\right)}^{{2}}}}},\ {\left|{x}\right|}{ < }{1}\)

Now, multiplying both sides by x leads to

\(\displaystyle{x}+{2}{x}^{{2}}+{3}{x}^{{3}}+\ldots={\sum_{{{n}={1}}}^{\infty}}{n}{x}^{{n}}={\frac{{{x}}}{{{\left({1}-{x}\right)}^{{2}}}}},\ {\left|{x}\right|}{ < }{1}\)

However, in this case \(\displaystyle{x}=\frac{{2}}{{3}}{ < }{1}\), so simply substitute \(\displaystyle{x}=\frac{{2}}{{3}}\) in formula above.

\(\displaystyle{1}+{x}+{x}^{{2}}+\ldots={\frac{{{1}}}{{{1}-{x}}}},\ {\left|{x}\right|}{ < }{1}\)

Moreover, power series can be differentiated term by term. So, differentiating both sides of the equation above we get

\(\displaystyle{1}+{2}{x}+{3}{x}^{{2}}+\ldots={\frac{{{1}}}{{{\left({1}-{x}\right)}^{{2}}}}},\ {\left|{x}\right|}{ < }{1}\)

Now, multiplying both sides by x leads to

\(\displaystyle{x}+{2}{x}^{{2}}+{3}{x}^{{3}}+\ldots={\sum_{{{n}={1}}}^{\infty}}{n}{x}^{{n}}={\frac{{{x}}}{{{\left({1}-{x}\right)}^{{2}}}}},\ {\left|{x}\right|}{ < }{1}\)

However, in this case \(\displaystyle{x}=\frac{{2}}{{3}}{ < }{1}\), so simply substitute \(\displaystyle{x}=\frac{{2}}{{3}}\) in formula above.