Question

Find the sum of the series 64.\(1 - ln\;2 + \frac{{{{(ln\;2)}^2}}}{{2!}} + \frac{{{{(ln\;2)}^3}}}{{3!}} + ......\)

Short answer

\(1 - \ln 2 + \frac{{{{(\ln 2)}^2}}}{{2!}} - \frac{{{{(\ln 2)}^3}}}{{3!}} + ...... = \frac{1}{2}\)

Step-by-step solution

Step 1: Form the series with help of examining the numerator and denominator.

In this problem we will first need to come up with a general term for the series. Notice that the numerators of all terms are just multiples of \(\ln 2\) - therefore, the general term will be \({(\ln 2)^n}\). Then notice that the denominator are factorials. That’s why the denominator in the general term will be \(n!\) . Finally the sign keeps changing, so the general term will have \({( - 1)^n}\) and we have the series:

\(1 - \ln 2 + \frac{{{{(\ln 2)}^2}}}{{2!}} - \frac{{{{(\ln 2)}^3}}}{{3!}} + ..... = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\frac{{{{(\ln 2)}^n}}}{{n!}}} \)

Step 2: Recognize familiar power series

Now we are trying to recognize a familiar power series similar to this one. Since there is an \(n!\) in the denominator, this ought to remind us of the power series for \({e^x}\)

\({e^x} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} \)

Step 3: Examination

Using this, we can now find the sum:

\(\begin{array}{l}\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{{{(\ln 2)}^n}}}{{n!}}} = \sum\limits_{n = 0}^\infty {\frac{{{{( - \ln 2)}^n}}}{{n!}} = {e^{ - \ln 2}}} \\\sum\limits_{n = 0}^\infty {{{( - 1)}^n}\frac{{{{(\ln 2)}^n}}}{{n!}} = {e^{\ln{\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 2}}\right.\\}\!\lower0.7ex\hbox{$2$}}}} = \frac{1}{2}} \end{array}\)