JEE maths#20

$\begin{aligned} a_1 & = \max (f(x)) \\ & = \max \left(\frac 1{2+\lfloor \sin x \rfloor}\right) & \small \color{#3D99F6} f(x) \text{ is maximum when } \lfloor \sin x \rfloor = -1 \text{, the minimum.} \\ & = \frac 1{2-1} = 1 \end{aligned}$

From:

$\begin{aligned} a_{n+1} & = a_n + \frac {(-1)^{n+2}}{n+1} \\ a_2 & = a_1 - \frac {(-1)^{1+2}}{1+1} \\ & = 1 - \frac 12 \\ a_3 & = 1 - \frac 12 + \frac 13 \\ a_4 & = 1 - \frac 12 + \frac 13 - \frac 14 \\ \cdots & = \cdots \\ \implies a_n & = 1 - \frac 12 + \frac 13 - \cdots + \frac {(-1)^{n+1}}n \\ \lim_{n \to \infty} a_n & = \boxed{\ln 2} \end{aligned}$

Here from the above function $\large f(x) = \frac{1}{ 2+ \lfloor \sin x\rfloor}$

we know that $f(x)$ will be maximum when $2+ \lfloor \sin x\rfloor$ will be minimum.

So $\lfloor \sin x \rfloor$ has minimum value as -1. So $max(f(x) = 1$ .

So $a_1=1$ , Now from the problem as stated we get $a_2=\dfrac{1}{2}$ , $a_3 = \frac{5}{6}$ , $a_4 = \frac{7}{12}$ and so on.

Hence we see that the series is like $1$ , $\frac{1}{2}$ , $\frac{5}{6}$ , $\frac{7}{12} \cdots$ .

Now see that each common difference in the above series is like $1, \frac{-1}{2} , \frac{1}{3} , \frac{-1}{4} ,\cdots$ .

So , We can say that $a_n=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...$ .

Now $\displaystyle{\lim_{n \rightarrow \infty} a_n} = \displaystyle{\lim_{n \rightarrow \infty} 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...}$ .

So we know by Taylor Series that $\ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3}+\cdots \infty$ .

putting $x=1$ , we get, $\ln 2 = \displaystyle{\lim_{n \rightarrow \infty} a_n}$ .

Hence answer is $\boxed{\ln2}$