\(\text{Given expression can be written as}\)
\(\displaystyle \Rightarrow \left(\frac{1}{2}-\frac{1}{3} \right)+\left(\frac{1}{4}-\frac{1}{5} \right)+\left(\frac{1}{6}-\frac{1}{7} \right)+\cdots\)
\(\displaystyle \Rightarrow \frac{1}{2}-\frac{1}{3} +\frac{1}{4}-\frac{1}{5} +\frac{1}{6}-\frac{1}{7} +\cdots \cdots (1)\)
\(\text{We know that}\)
\(\displaystyle \ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3} -\frac{x^4}{4} +\cdots\)
\(\text{So from this we can say }\)
\(\displaystyle \ln(1+1)=\ln(2) = 1-\frac{1}{2}+\frac{1}{3} -\frac{1}{4} +\cdots\)
\(\text{If you look carefully (1) is noting but some adjustemnt of above series i.e}\)
\(\displaystyle \ln(2) = 1-\left(\frac{1}{2}-\frac{1}{3} +\frac{1}{4} +\cdots \right) \cdots \cdots (2)\)
\(\text{So form (2)}\)
\(\displaystyle 1-\ln(2)=\left(\frac{1}{2}-\frac{1}{3} +\frac{1}{4} +\cdots \right) \cdots \cdots (2)\)
\(\text{ANSWER:}\;\;\;\; 1- \ln(2)\)
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