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Evaluate $\displaystyle \sum_{n \,=\, 2}^{127} \log_{2} \Bigg(1+\dfrac{1}{n}\Bigg)$

$n$ is a literal number and it represents the numbers from $2$ to $127$. A logarithmic term whose number is sum of $1$ and reciprocal of $n$ and its base $2$. The sum of the binary logarithm terms for the $n$ values start from $2$ to $127$ is symbolically represented by the uppercase sigma symbol.

$\displaystyle \sum_{n \,=\, 2}^{127} \log_{2} \Bigg(1+\dfrac{1}{n}\Bigg)$

The value of the function is required to evaluate in this logarithm problem.



Substitute $n = 2$ in the logarithmic term and repeat the procedure till $n = 127$ but add all of them to represent the functionality of the uppercase sigma symbol.

$=\,\,\,$ $\log_{2} \Bigg(1+\dfrac{1}{2}\Bigg)$ $+$ $\log_{2} \Bigg(1+\dfrac{1}{3}\Bigg)$ $+$ $\log_{2} \Bigg(1+\dfrac{1}{4}\Bigg)$ $+$ $\cdots$ $+$ $\log_{2} \Bigg(1+\dfrac{1}{127}\Bigg)$

$=\,\,\,$ $\log_{2} \Bigg(\dfrac{2+1}{2}\Bigg)$ $+$ $\log_{2} \Bigg(\dfrac{3+1}{3}\Bigg)$ $+$ $\log_{2} \Bigg(\dfrac{4+1}{4}\Bigg)$ $+$ $\cdots$ $+$ $\log_{2} \Bigg(\dfrac{127+1}{127}\Bigg)$

$=\,\,\,$ $\log_{2} \Bigg(\dfrac{3}{2}\Bigg)$ $+$ $\log_{2} \Bigg(\dfrac{4}{3}\Bigg)$ $+$ $\log_{2} \Bigg(\dfrac{5}{4}\Bigg)$ $+$ $\cdots$ $+$ $\log_{2} \Bigg(\dfrac{128}{127}\Bigg)$


Merging logarithmic terms

The sum of all logarithmic terms can be merged as a single logarithmic term by using product rule of logarithms.

$=\,\,\,$ $\log_{2} \Bigg[\Bigg(\dfrac{3}{2}\Bigg) \Bigg(\dfrac{4}{3}\Bigg) \Bigg(\dfrac{5}{4}\Bigg) \cdots \Bigg(\dfrac{128}{127}\Bigg)\Bigg]$

$=\,\,\,$ $\log_{2} \Bigg(\dfrac{3 \times 4 \times 5 \times \cdots \times 128}{2 \times 3 \times 4 \times \cdots \times 127}\Bigg)$

$=\,\,\,$ $\require{cancel} \log_{2} \Bigg(\dfrac{\cancel{3} \times \cancel{4} \times \cancel{5} \times \cdots \times 128}{2 \times \cancel{3} \times \cancel{4} \times \cdots \times \cancel{127}}\Bigg)$

$=\,\,\,$ $\log_{2} \Bigg(\dfrac{128}{2}\Bigg)$

$=\,\,\,$ $\require{cancel} \log_{2} \Bigg(\dfrac{\cancel{128}}{\cancel{2}}\Bigg)$

$=\,\,\,$ $\log_{2} 64$


Finding the value

The base of logarithmic term is $2$. So, express the number in exponential notation on the basis of number $2$.

$=\,\,\,$ $\log_{2} 2^6$

It can be simplified by the power rule of logarithms.

$=\,\,\,$ $6 \times \log_{2} 2$

The logarithm of the base is one as per the logarithm of base rule.

$=\,\,\,$ $6 \times 1$

$=\,\,\,$ $6$

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