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# The sum of series ${\log _4}2 - {\log _8}2 + {\log _{16}}2$…………..isA. ${e^2}$B. ${\log _e}2$C. ${\log _e}3 - 2$D. $1 - {\log _e}2$

Last updated date: 18th Mar 2023
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Hint: First simplify the sum using various logarithmic properties & then use the Maclaurin’s series to find the solution.

Here we have to find the sum of series of ${\log _4}2 - {\log _8}2 + {\log _{16}}2……..$
This can be written as ${\log _{{2^2}}}2 - {\log _{{2^3}}}2 + {\log _{{2^4}}}2……….$
Now we will be using the property of logarithm, ${\log _b}a = \dfrac{{\ln a}}{{\ln b}}$ and ${\log _{{b^p}}}a = \dfrac{1}{p}{\log _b}a$
So using the above property we can write our series as
$\dfrac{1}{2}\dfrac{{\ln 2}}{{\ln 2}} - \dfrac{1}{3}\dfrac{{\ln 2}}{{\ln 2}} + \dfrac{1}{4}\dfrac{{\ln 2}}{{\ln 2}}................\infty$

$\ln 2$ gets cancelled from the numerator and denominator, so we get
$\Rightarrow \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4}............\infty$
Let’s take negative common from the above sum, we get
$- \left[ {\dfrac{{ - 1}}{2} + \dfrac{1}{3} - \dfrac{1}{4}................\infty } \right]$………………………….. (1)
Now the Maclaurin’s series expansion for $\ln (1 + x) = x - {\dfrac{x}{2}^2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ..................$
Let’s add and subtract 1 from equation (1)
$- \left[ {1 + \dfrac{{ - 1}}{2} + \dfrac{1}{3} - \dfrac{1}{4} - 1................\infty } \right]$
Clearly from $1 - \dfrac{1}{2} + \dfrac{1}{3}..........\infty$ leaving $-1$ forms Maclaurin’s series expansion for $\ln (1 + x)$ at $x = 1$
So we can write above as
$- \left[ {\ln (1 + 1) - 1} \right]$
This is nothing but
$- \left[ {\ln (2) - 1} \right]$
$1 - \ln (2)$
Using $\ln a = {\log _e}a$ we can write the above value as
$1 - {\log _e}2$