Answer

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Hint: First simplify the sum using various logarithmic properties & then use the Maclaurin’s series to find the solution.

Complete step-by-step answer:

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]$

Multiplying the negative sign inside, we get

$1 - \ln (2)$

Using $\ln a = {\log _e}a$ we can write the above value as

$1 - {\log _e}2$

So option (D) is correct.

Note: Whenever we come across such problems, the key point is to figure out which series expansion we are dealing with and to retrieve back the function from this series, important logarithm properties are also advised to be remembered as it helps solving such problems.

Complete step-by-step answer:

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]$

Multiplying the negative sign inside, we get

$1 - \ln (2)$

Using $\ln a = {\log _e}a$ we can write the above value as

$1 - {\log _e}2$

So option (D) is correct.

Note: Whenever we come across such problems, the key point is to figure out which series expansion we are dealing with and to retrieve back the function from this series, important logarithm properties are also advised to be remembered as it helps solving such problems.

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