$1+\dfrac{1}{4} \left(\dfrac{1}{2}+\dfrac{1}{3}\right) + \frac{1}{4^2} \left(\dfrac{1}{4}+\dfrac{1}{5}\right) + \dfrac{1}{4^3} \left(\dfrac{1}{6}+\dfrac{1}{7}\right)+\cdots$

If the value of the series above is equal to $\ln \sqrt A$, find $A$.

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