Question

Which of the following options is/are correct, if

S : If $a_{n-1} = \frac{a_n-n}{n} \forall n \in N -\{1\}, a_1 = 1$, then $\lim_{n\to\infty} \prod_{r=1}^{n} (1 + \frac{1}{a_r}) = e$.

T : $\lim_{n\to\infty} (\frac{n^2+n+1}{n+1}-(an+b)) = 4 \Rightarrow 7a-2b=15$.

• A. S is True and T is False
• B. S is False and T is True
• C. Both S and T are True
• D. Both S and T are Flase

Answer 1

Answer: (C)

Both S and T are True

Answer 2

Statement S: The given recurrence relation is $a_{n-1} = \frac{a_n-n}{n}$ for $n \in N -\{1\}$, which means $n \ge 2$. This can be rewritten as $n a_{n-1} = a_n - n$, so $a_n = n a_{n-1} + n = n(a_{n-1} + 1)$ for $n \ge 2$. Given $a_1 = 1$. Let's check if the relation $a_n = n(a_{n-1} + 1)$ holds for $n=1$. If we substitute $n=1$, we get $a_1 = 1(a_0+1)$, which involves $a_0$. The recurrence is given for $n \ge 2$. However, let's rewrite the relation as $a_{n-1} + 1 = \frac{a_n}{n}$ for $n \ge 2$. For $n=2$: $a_1 + 1 = \frac{a_2}{2}$. Since $a_1=1$, $1+1 = \frac{a_2}{2} \Rightarrow a_2 = 4$. For $n=3$: $a_2 + 1 = \frac{a_3}{3}$. Since $a_2=4$, $4+1 = \frac{a_3}{3} \Rightarrow a_3 = 15$. For $n=4$: $a_3 + 1 = \frac{a_4}{4}$. Since $a_3=15$, $15+1 = \frac{a_4}{4} \Rightarrow a_4 = 64$. The relation $a_{r-1} + 1 = \frac{a_r}{r}$ holds for $r \ge 2. This is equivalent to$a_r + 1 = \frac{a_{r+1}}{r+1}$for$r \ge 1$.

We need to evaluate the limit of the product $P_n = \prod_{r=1}^{n} (1 + \frac{1}{a_r})$ as $n \to \infty$. $P_n = \prod_{r=1}^{n} \frac{a_r+1}{a_r}$. Using the relation $a_r+1 = \frac{a_{r+1}}{r+1}$ for $r \ge 1$, we have $\frac{a_r+1}{a_r} = \frac{a_{r+1}/(r+1)}{a_r} = \frac{a_{r+1}}{(r+1)a_r}$. So, $P_n = \prod_{r=1}^{n} \frac{a_{r+1}}{(r+1)a_r} = \frac{a_2}{2a_1} \cdot \frac{a_3}{3a_2} \cdot \frac{a_4}{4a_3} \cdot \dots \cdot \frac{a_{n+1}}{(n+1)a_n}$. This is a telescoping product: $P_n = \frac{a_2 \cdot a_3 \cdot \dots \cdot a_{n+1}}{(2 \cdot 3 \cdot \dots \cdot (n+1)) \cdot (a_1 \cdot a_2 \cdot \dots \cdot a_n)} = \frac{a_{n+1}}{(n+1)! a_1}$. Since $a_1 = 1$, $P_n = \frac{a_{n+1}}{(n+1)!}$.

From the relation $a_n = n(a_{n-1} + 1)$ for $n \ge 2$, we have $\frac{a_n}{n!} = \frac{n(a_{n-1}+1)}{n!} = \frac{a_{n-1}+1}{(n-1)!}$ for $n \ge 2$. Let $b_n = \frac{a_n}{n!}$. Then $b_n = \frac{a_{n-1}}{(n-1)!} + \frac{1}{(n-1)!} = b_{n-1} + \frac{1}{(n-1)!}$ for $n \ge 2$. This is a recurrence relation for $b_n$. $b_2 = b_1 + \frac{1}{1!}$ $b_3 = b_2 + \frac{1}{2!} = b_1 + \frac{1}{1!} + \frac{1}{2!}$ $b_n = b_1 + \sum_{k=1}^{n-1} \frac{1}{k!}$ for $n \ge 2$. $b_1 = \frac{a_1}{1!} = \frac{1}{1} = 1$. So, $b_n = 1 + \sum_{k=1}^{n-1} \frac{1}{k!} = \sum_{k=0}^{n-1} \frac{1}{k!}$ for $n \ge 2$. For $n=1$, $b_1 = \frac{a_1}{1!} = 1$, and $\sum_{k=0}^{1-1} \frac{1}{k!} = \frac{1}{0!} = 1$. So the formula $b_n = \sum_{k=0}^{n-1} \frac{1}{k!}$ holds for $n \ge 1$.

We have $P_n = \frac{a_{n+1}}{(n+1)!} = b_{n+1}$. So, $P_n = \sum_{k=0}^{(n+1)-1} \frac{1}{k!} = \sum_{k=0}^{n} \frac{1}{k!}$. We need to find $\lim_{n\to\infty} P_n = \lim_{n\to\infty} \sum_{k=0}^{n} \frac{1}{k!}$. This is the definition of the exponential constant $e$. $\lim_{n\to\infty} \sum_{k=0}^{n} \frac{1}{k!} = e$. So, $\lim_{n\to\infty} \prod_{r=1}^{n} (1 + \frac{1}{a_r}) = e$. Statement S is True.

Statement T: We are given $\lim_{n\to\infty} (\frac{n^2+n+1}{n+1}-(an+b)) = 4$. Let's simplify the expression inside the limit: $\frac{n^2+n+1}{n+1}-(an+b) = \frac{n^2+n+1 - (an+b)(n+1)}{n+1}$ $= \frac{n^2+n+1 - (an^2 + an + bn + b)}{n+1}$ $= \frac{(1-a)n^2 + (1-a-b)n + (1-b)}{n+1}$. For the limit as $n \to \infty$ to be a finite value (4), the degree of the numerator must be less than or equal to the degree of the denominator. The degree of the denominator is 1. The degree of the numerator is 2. For the limit to be finite, the coefficient of $n^2$ in the numerator must be zero. $1-a = 0 \Rightarrow a = 1$. Substitute $a=1$ into the expression: $\frac{(1-1)n^2 + (1-1-b)n + (1-b)}{n+1} = \frac{-bn + (1-b)}{n+1}$. Now, evaluate the limit: $\lim_{n\to\infty} \frac{-bn + (1-b)}{n+1}$. Divide numerator and denominator by $n$: $\lim_{n\to\infty} \frac{-b + \frac{1-b}{n}}{1 + \frac{1}{n}} = \frac{-b + 0}{1 + 0} = -b$. We are given that this limit is 4. So, $-b = 4 \Rightarrow b = -4$. We found $a=1$ and $b=-4$. The statement T claims that if the limit is 4, then $7a-2b=15$. Let's substitute the values of $a$ and $b$ we found into the expression $7a-2b$: $7(1) - 2(-4) = 7 + 8 = 15$. The condition $7a-2b=15$ is satisfied with $a=1$ and $b=-4$. Therefore, Statement T is True.

Both Statement S and Statement T are True.