LimitshardFree

Limit of a bracketed product raised to n and divided by n^(n^2) | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
lim⁡n→∞1nn2[(n+1)(n+12)(n+122)⋯(n+12n−1)]n\displaystyle\lim_{n\to\infty}\frac{1}{n^{n^2}}\left[\left(n+1\right)\left(n+\frac12\right) \left(n+\frac{1}{2^2}\right)\cdots\left(n+\frac{1}{2^{n-1}}\right)\right]^{n} is(are) 'aa', then
A[a]=5\left[a\right] = 5
B[a]=7\left[a\right] = 7correct
C(a)=6\left(a\right) = 6
D(a)=8\left(a\right) = 8correct
Solution
Step 1: Taking a factor nn out of each of the nn brackets,
∏k=0n−1(n+12k)=nn∏k=0n−1(1+12kn)\prod_{k=0}^{n-1}\left(n+\frac{1}{2^{k}}\right) = n^{n}\prod_{k=0}^{n-1}\left(1+\frac{1}{2^{k}n}\right)
[ ⋅ ]n=nn2[∏k=0n−1(1+12kn)]n\left[\,\cdot\,\right]^{n} = n^{n^2}\left[\prod_{k=0}^{n-1}\left(1+\frac{1}{2^{k}n}\right)\right]^{n}
 so the prefactor n−n2n^{-n^2} cancels and
a=lim⁡n→∞[∏k=0n−1(1+12kn)]na = \lim_{n\to\infty}\left[\prod_{k=0}^{n-1}\left(1+\frac{1}{2^{k}n}\right)\right]^{n}
 Step 2: Taking logarithms,
ln⁡a=lim⁡n→∞ n∑k=0n−1ln⁡(1+12kn)\ln a = \lim_{n\to\infty}\ n\sum_{k=0}^{n-1}\ln\left(1+\frac{1}{2^{k}n}\right)
 Step 3: For u>0u>0, ddu[ln⁡(1+u)−u+u22]=u21+u>0\dfrac{d}{du}\left[\ln\left(1+u\right)-u+\dfrac{u^2}{2}\right] = \dfrac{u^2}{1+u}>0, and the bracket is 00 at u=0u = 0, so
0≤u−ln⁡(1+u)≤u22(u>0)0 \le u-\ln\left(1+u\right) \le \frac{u^2}{2}\qquad\left(u>0\right)
 Putting u=12knu = \dfrac{1}{2^{k}n} and adding over kk,
0≤n∑k=0n−1[12kn−ln⁡(1+12kn)]≤n2∑k=0n−114kn2<12n⋅43=23n→00 \le n\sum_{k=0}^{n-1}\left[\frac{1}{2^{k}n}-\ln\left(1+\frac{1}{2^{k}n}\right)\right] \le \frac{n}{2}\sum_{k=0}^{n-1}\frac{1}{4^{k}n^{2}} < \frac{1}{2n}\cdot\frac43 = \frac{2}{3n} \to 0
 so replacing ln⁡(1+u)\ln\left(1+u\right) by uu does not change the limit:
ln⁡a=lim⁡n→∞ n∑k=0n−112kn=lim⁡n→∞∑k=0n−112k\ln a = \lim_{n\to\infty}\ n\sum_{k=0}^{n-1}\frac{1}{2^{k}n} = \lim_{n\to\infty}\sum_{k=0}^{n-1}\frac{1}{2^{k}}
 Step 4:
∑k=0∞12k=11−12=2⟹ln⁡a=2,a=e2\sum_{k=0}^{\infty}\frac{1}{2^{k}} = \frac{1}{1-\tfrac12} = 2 \quad\Longrightarrow\quad \ln a = 2 , \quad a = e^{2}
 Step 5: 2.7<e<2.82.7<e<2.8, so
7.29<e2<7.847.29 < e^{2} < 7.84
e2=7.389…,7<e2<8e^{2} = 7.389\ldots ,\qquad 7 < e^{2} < 8
 Step 6: Greatest integer [a]=7\left[a\right] = 7, so (2) is correct and (1) is false. Least integer (a)=8\left(a\right) = 8, so (4) is correct and (3) is false. Answer: (2) and (4)
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