Let the sequence of real numbers satisfies the recurrence relation bn+1=(1/3)(2 bn+(125/bn2)), bn ≠ 0 . Then find displaystyle lim n arrow ∞ bn

Question

Let the sequence < bn > of real numbers satisfies the recurrence relation bn+1=(1/3)(2 bn+(125/bn2)), bn ≠ 0 . Then find displaystyle lim n arrow ∞ bn (A) 10 (B) 15 (C) 5 (D) 25

Tardigrade Admin · Accepted Answer

Let displaystyle lim n arrow ∞ bn=b Now, bn+1=(1/3)(2 bn+(125/bn2)) or displaystyle lim n arrow ∞ bn+1=(1/3)(2 displaystyle lim n arrow ∞ bn+(125/ displaystyle lim n arrow ∞ bn2)) or b=(1/3)(2 b+(125/b2)) ⇒ (b/3)=(125/3 b2) ⇒ b3=125 or b=5.

Q. Let the sequence <bn​> of real numbers satisfies the recurrence relation bn+1​=31​(2bn​+bn2​125​),bn​=0. Then find n→∞lim​bn​

10

15

5

25

Let n→∞lim​bn​=b Now, bn+1​=31​(2bn​+bn2​125​) or n→∞lim​bn+1​=31​⎝⎛​2n→∞lim​bn​+n→∞lim​bn2​125​⎠⎞​ or b=31​(2b+b2125​) ⇒3b​=3b2125​ ⇒b3=125 or b=5.

Q. Let the sequence $< b_{n} >$ of real numbers satisfies the recurrence relation $b_{n + 1} = \frac{1}{3} (2 b_{n} + \frac{125}{b _{n}^{2}}), b_{n} \neq = 0.$ Then find $n \to \infty lim b_{n}$