3.0 Logarithmic Series

  Exponential and Logarithmic Series

If $\left| x \right| < 1$ then, $$ {\log _e}(1 + x) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + ...\infty $$ or $$ = {\log _e}(1 + x) = \sum\limits_{r = 1}^\infty {{{( - 1)}^{r - 1}}\frac{{{x^r}}}{r}} $$

Similarly,$$ {\log _e}(1 - x) = - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} - ...\infty $$ or $$ = {\log _e}(1 - x) = \sum\limits_{r = 1}^\infty {{{( - 1)}}\frac{{{x^r}}}{r}} $$

Proof: We know,$$ = {\log _e}(1 + x) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + ...\infty $$ Put $x \to \left( { - x} \right)$. So, the above equation becomes, $$ = {\log _e}(1 - x) = - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} - ...\infty $$

Question 6. Prove that, $$ = \frac{{(p - 1) - \frac{1}{2}{{(p - 1)}^2} + \frac{1}{3}{{\left( {p - 1} \right)}^3} - ...\infty }}{{(q - 1) - \frac{1}{2}{{(q - 1)}^2} + \frac{1}{3}{{\left( {q - 1} \right)}^3} - ...\infty }} = {\log _q}p$$

Solution: We know, $$ = {\log _e}(1 + x) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + ...\infty $$ So, the numerator can be writtern as, $$\begin{equation} \begin{aligned} = (p - 1) - \frac{1}{2}{(p - 1)^2} + \frac{1}{3}{\left( {p - 1} \right)^3} - ...\infty = \log \left( {1 + \left\{ {p - 1} \right\}} \right) \\ = (p - 1) - \frac{1}{2}{(p - 1)^2} + \frac{1}{3}{\left( {p - 1} \right)^3} - ...\infty = \log p \\\end{aligned} \end{equation} $$ Similarly, $$\begin{equation} \begin{aligned} = (q - 1) - \frac{1}{2}{(q - 1)^2} + \frac{1}{3}{\left( {q - 1} \right)^3} - ...\infty = \log \left( {1 + \left\{ {q - 1} \right\}} \right) \\ = (q - 1) - \frac{1}{2}{(q - 1)^2} + \frac{1}{3}{\left( {q - 1} \right)^3} - ...\infty = \log q \\\end{aligned} \end{equation} $$ So, after solving numerator and denominator we get, $$ = \frac{{\log p}}{{\log q}} = {\log _q}p$$

Question 7. Prove that the series, $$ \frac{1}{{n + 1}} + \frac{1}{{2{{(n + 1)}^2}}} + \frac{1}{{3{{(n + 1)}^3}}} + ...$$ has the same sum as the series $$ = \frac{1}{n} - \frac{1}{{2{n^2}}} + \frac{1}{{3{n^3}}} - ...$$

Solution: Let, $\left( {\frac{1}{{n + 1}}} \right) = x$

So, the above equation becomes, $$ x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} + ...\infty $$ We know, $$\begin{equation} \begin{aligned} = {\log _e}(1 - x) = - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} - ...\infty \\ = {\log _e}(1 - x) = - \left( {x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} + \frac{{{x^4}}}{4} + ...\infty } \right) \\\end{aligned} \end{equation} $$ On comparing the above equations we get, $$\begin{equation} \begin{aligned} = \frac{1}{{n + 1}} + \frac{1}{{2{{(n + 1)}^2}}} + \frac{1}{{3{{(n + 1)}^3}}} + ... = - {\log _e}\left( {1 - \frac{1}{{n + 1}}} \right) \\ = - {\log _e}\left( {\frac{{n + 1 - 1}}{{n + 1}}} \right) \\ = {\log _e}{\left( {\frac{n}{{n + 1}}} \right)^{ - 1}} \\ = {\log _e}\left( {\frac{{n + 1}}{n}} \right) \\\end{aligned} \end{equation} $$ Similarly for, $$ = \frac{1}{n} - \frac{1}{{2{n^2}}} + \frac{1}{{3{n^3}}} - ...$$ Let $\frac{1}{n} = a$

The above equation becomes, $$ = a - \frac{{{a^2}}}{2} + \frac{{{a^3}}}{3} - ...$$ We know, $$ = {\log _e}(1 - x) = - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} - ...\infty $$ On comparing the above equations we get, $$ = a - \frac{{{a^2}}}{2} + \frac{{{a^3}}}{3} - ...\infty = {\log _e}(1 + a)$$ For $a = \frac{1}{n}$ we get, $$\begin{equation} \begin{aligned} = {\log _e}\left( {1 + \frac{1}{n}} \right) \\ = {\log _e}\left( {\frac{{n + 1}}{n}} \right) \\\end{aligned} \end{equation} $$