Question:

Step 1: Rewrite the general term using a substitution. The general term of the series is $a_r = \dfrac{\sqrt{t^{2r-3}}(1-t)}{(\sqrt{t^{2r-1}}+1)(\sqrt{t^{2r-3}}+1)}$. Let $u_r = \sqrt{t^{2r-3}} = t^{(2r-3)/2}$. Then $u_{r+1} = t^{(2(r+1)-3)/2} = t^{(2r-1)/2}$. Observe that $t u_r = t \cdot t^{(2r-3)/2} = t^{1 + (2r-3)/2} = t^{(2+2r-3)/2} = t^{(2r-1)/2}$. Thus, $\sqrt{t^{2r-1}} = t u_r$. The general term can be rewritten as $a_r = \dfrac{u_r(1-t)}{(tu_r+1)(u_r+1)}$. Step 2: Express the general term as a telescoping difference. Consider the difference of two terms: $$ \frac{1}{tu_r+1} - \frac{1}{u_r+1} = \frac{(u_r+1) - (tu_r+1)}{(tu_r+1)(u_r+1)} = \frac{u_r - tu_r}{(tu_r+1)(u_r+1)} = \frac{u_r(1-t)}{(tu_r+1)(u_r+1)} $$ This matches the general term $a_r$. Substituting back $u_r = \sqrt{t^{2r-3}}$ and $tu_r = \sqrt{t^{2r-1}}$, we have: $$ a_r = \frac{1}{\sqrt{t^{2r-1}}+1} - \frac{1}{\sqrt{t^{2r-3}}+1} $$ Step 3: Sum the telescoping series from $r=2$ to $n$. Let $S_n$ be the partial sum: $$ S_n = \sum_{r=2}^{n} \left(\frac{1}{\sqrt{t^{2r-1}}+1} - \frac{1}{\sqrt{t^{2r-3}}+1}\right) $$ Expanding the sum, we observe cancellation of intermediate terms: $$ S_n = \left(\frac{1}{\sqrt{t^3}+1} - \frac{1}{\sqrt{t}+1}\right) + \left(\frac{1}{\sqrt{t^5}+1} - \frac{1}{\sqrt{t^3}+1}\right) + \dots + \left(\frac{1}{\sqrt{t^{2n-1}}+1} - \frac{1}{\sqrt{t^{2n-3}}+1}\right) $$ The sum simplifies to: $$ S_n = \frac{1}{\sqrt{t^{2n-1}}+1} - \frac{1}{\sqrt{t}+1} $$ Step 4: Take the limit as $n \to \infty$ to find $P(t)$. For $0 < t < 1$, as $n \to \infty$, $t^{2n-1} \to 0$. Therefore, $\lim_{n \to \infty} \dfrac{1}{\sqrt{t^{2n-1}}+1} = \dfrac{1}{\sqrt{0}+1} = 1$. So, $P(t) = \lim_{n \to \infty} S_n = 1 - \dfrac{1}{\sqrt{t}+1}$. This expression can be simplified: $$ P(t) = \frac{(\sqrt{t}+1)-1}{\sqrt{t}+1} = \frac{\sqrt{t}}{\sqrt{t}+1} $$ Step 5: Evaluate $P\left(\dfrac{1}{2}\right)$. Substitute $t = \dfrac{1}{2}$ into the expression for $P(t)$: $$ P\left(\frac{1}{2}\right) = \frac{\sqrt{1/2}}{\sqrt{1/2}+1} = \frac{1/\sqrt{2}}{1/\sqrt{2}+1} $$ To simplify, multiply the numerator and denominator by $\sqrt{2}$: $$ P\left(\frac{1}{2}\right) = \frac{1}{1+\sqrt{2}} $$ Rationalize the denominator by multiplying by the conjugate $\sqrt{2}-1$: $$ P\left(\frac{1}{2}\right) = \frac{1}{1+\sqrt{2}} \cdot \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{\sqrt{2}-1}{(\sqrt{2})^2-1^2} = \frac{\sqrt{2}-1}{2-1} = \sqrt{2}-1 $$ Step 6: Determine the interval for $P\left(\dfrac{1}{2}\right)$. The numerical value of $P\left(\dfrac{1}{2}\right) = \sqrt{2}-1 \approx 1.4142 - 1 = 0.4142$. We compare this value with common trigonometric values: $\sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2} = 0.5$. Since $0 < 0.4142 < 0.5$, it follows that $P\left(\dfrac{1}{2}\right)$ lies in the interval $\left(0, \sin\dfrac{\pi}{6}\right)$. Additionally, $\tan\left(\dfrac{\pi}{4}\right) = 1$. Since $0 < 0.4142 < 1$, it follows that $P\left(\dfrac{1}{2}\right)$ also lies in the interval $\left(0, \tan\dfrac{\pi}{4}\right)$.