Question

Determine the sum of the series.

\(\sum\limits_{n = 1}^\infty {\frac{{{4^n}}}{{n{5^n}}}} \)

Short answer

The sum of the series \(\sum\limits_{n = 1}^\infty {\frac{{{4^n}}}{{n{5^n}}}} \) is approximately\(1.61\).

Step-by-step solution

Term by term integration of power series

If \({(x - a)^n}\) is differentiable for all\(x\), then according to term-by-term integration of power series, we have

\(\sum\limits_{n = 0}^\infty {{c_n}} \frac{{{{(x - a)}^{n + 1}}}}{{n + 1}} = \int {\left( {\sum\limits_{n = 0}^\infty {{c_n}} {{(x - a)}^n}} \right)} dx\)

As\({x^n}\)is differentiable for all\(x\), therefore we have

\(\begin{aligned}S &= \sum\limits_{n = 0}^\infty {\frac{{{x^{n + 1}}}}{{n + 1}}} \\S &= \int {\left( {\sum\limits_{n = 0}^\infty {{x^n}} } \right)} dx\\S &= \int {\left( {\frac{1}{{1 - x}}} \right)} dx\end{aligned}\)

Calculate the sum of the given series

As given, the sum of the series\(\sum\limits_{n = 1}^\infty {\frac{{{4^n}}}{{n{5^n}}}} \)

Compute the sum of the series \(\sum\limits_{n = 1}^\infty {\frac{{{4^n}}}{{n{5^n}}}} \) as follows

\(\begin{aligned}\sum\limits_{n = 1}^\infty {\frac{{{4^n}}}{{n{5^n}}}} &= \sum\limits_{n = 1}^\infty {\left( {\frac{1}{n}} \right)} \frac{{{4^n}}}{{{5^n}}}\\\sum\limits_{n = 1}^\infty {\frac{{{4^n}}}{{n{5^n}}}} &= \sum\limits_{n = 1}^\infty {\frac{1}{n}} {\left( {\frac{4}{5}} \right)^n}\end{aligned}\)

Let \(x = \frac{4}{5}\), that implies

\(\sum\limits_{n = 1}^\infty {\frac{1}{n}} {x^n} = \sum\limits_{n = 0}^\infty {\frac{{{x^{n + 1}}}}{{n + 1}}} \)

Integrate \(\sum\limits_{n = 0}^\infty  {{x^n}} \)

Integrate \(\sum\limits_{n = 0}^\infty {{x^n}} \) as follows

\(\int {\sum\limits_{n = 0}^\infty {{x^n}} } dx = \int {\frac{1}{{1 - x}}} dx\)

Let,\(u = 1 - x\),

Then, \(du = - dx\)

\(\begin{aligned}\int {\sum\limits_{n = 0}^\infty {{x^n}} } dx &= \int {\frac{1}{{1 - x}}} dx\\\int {\sum\limits_{n = 0}^\infty {{x^n}} } dx &= \int {\frac{1}{u}} ( - du)\\\int {\sum\limits_{n = 0}^\infty {{x^n}} } dx &= - \ln |u|\\\int {\sum\limits_{n = 0}^\infty {{x^n}} } dx &= - \ln |1 - x|\end{aligned}\)

On further simplification,

\(\begin{aligned}\int {\sum\limits_{n = 0}^\infty {{x^n}} } dx &= - \ln \left( {1 - \frac{4}{5}} \right)\\\int {\sum\limits_{n = 0}^\infty {{x^n}} } dx &= - \ln \left( {\frac{{5 - 4}}{5}} \right)\\\int {\sum\limits_{n = 0}^\infty {{x^n}} } dx &= - \ln \left( {\frac{1}{5}} \right)\\\int {\sum\limits_{n = 0}^\infty {{x^n}} } dx &= - 1.60943\\\int {\sum\limits_{n = 0}^\infty {{x^n}} } dx &\approx - 1.61\end{aligned}\)

Therefore, the sum of the series \(\sum\limits_{n = 1}^\infty {\frac{{{4^n}}}{{n{5^n}}}} \) is approximately\(1.61\).