Convergence proof and limit determination

The question asks to prove that a sequence converges (using monotone convergence, fixed point arguments, etc.) and/or to determine its limit, typically by solving f(ℓ) = ℓ.

isi-entrance 2012 Q17 View
Let $a_1 = 24^{1/3}$ and $a_{n+1} = (a_n + 24)^{1/3}$. Find the integer part of $a_{100}$.
isi-entrance 2012 Q19 View
Find $\lim_{n\to\infty}\left(1 + \dfrac{1}{n}\right)^n$.
isi-entrance 2020 Q4 View
Let a real-valued sequence $\left\{x_{n}\right\}_{n \geq 1}$ be such that
$$\lim_{n \rightarrow \infty} n x_{n} = 0$$
Find all possible real values of $t$ such that $\lim_{n \rightarrow \infty} x_{n}(\log n)^{t} = 0$.
jee-advanced 2024 Q5 4 marks View
Let $S$ be the set of all $( \alpha , \beta ) \in \mathbb { R } \times \mathbb { R }$ such that
$$\lim _ { x \rightarrow \infty } \frac { \sin \left( x ^ { 2 } \right) \left( \log _ { e } x \right) ^ { \alpha } \sin \left( \frac { 1 } { x ^ { 2 } } \right) } { x ^ { \alpha \beta } \left( \log _ { e } ( 1 + x ) \right) ^ { \beta } } = 0 .$$
Then which of the following is (are) correct?
(A) $( - 1,3 ) \in S$
(B) $( - 1,1 ) \in S$
(C) $( 1 , - 1 ) \in S$
(D) $( 1 , - 2 ) \in S$
jee-main 2021 Q61 View
The value of $4 + \frac { 1 } { 5 + \frac { 1 } { 4 + \frac { 1 } { 5 + \frac { 1 } { 4 + \ldots . . \infty } } } }$ is:
(1) $2 + \frac { 2 } { 5 } \sqrt { 30 }$
(2) $2 + \frac { 4 } { \sqrt { 5 } } \sqrt { 30 }$
(3) $4 + \frac { 4 } { \sqrt { 5 } } \sqrt { 30 }$
(4) $5 + \frac { 2 } { 5 } \sqrt { 30 }$
jee-main 2021 Q61 View
The value of $3 + \frac { 1 } { 4 + \frac { 1 } { 3 + \frac { 1 } { 4 + \frac { 1 } { 3 + \ldots . \infty } } } }$ is equal to
(1) $1.5 + \sqrt { 3 }$
(2) $2 + \sqrt { 3 }$
(3) $3 + 2 \sqrt { 3 }$
(4) $4 + \sqrt { 3 }$
jee-main 2022 Q68 View
Let $\beta = \lim _ { x \rightarrow 0 } \frac { \alpha x - \left( e ^ { 3 x } - 1 \right) } { \alpha x \left( e ^ { 3 x } - 1 \right) }$ for some $\alpha \in \mathbb { R }$. Then the value of $\alpha + \beta$ is:
(1) $\frac { 14 } { 5 }$
(2) $\frac { 3 } { 2 }$
(3) $\frac { 5 } { 2 }$
(4) $\frac { 1 } { 2 }$
jee-main 2022 Q66 View
If $\lim _ { n \rightarrow \infty } \left( \sqrt { n ^ { 2 } - n - 1 } + n\alpha + \beta \right) = 0$ then $8\alpha + \beta$ is equal to
(1) 4
(2) - 8
(3) - 4
(4) 8
jee-main 2022 Q89 View
If $\lim _ { n \rightarrow \infty } \frac { (n+1)^{k-1} } { n ^ { k + 1 } } \left[ (nk+1) + (nk+2) + \ldots + (nk+n) \right] = 33 \cdot \lim _ { n \rightarrow \infty } \frac { 1 } { n ^ { k + 1 } } \cdot \left( 1 ^ { k } + 2 ^ { k } + 3 ^ { k } + \ldots + n ^ { k } \right)$, then the integral value of $k$ is equal to $\_\_\_\_$.
jee-main 2023 Q65 View
$\lim _ { t \rightarrow 0 } 1 ^ { \frac { 1 } { \sin ^ { 2 } t } } + 2 ^ { \frac { 1 } { \sin ^ { 2 } t } } + 3 ^ { \frac { 1 } { \sin ^ { 2 } t } } \ldots \ldots n ^ { \frac { 1 } { \sin ^ { 2 } t } } \sin ^ { 2 } t$ is equal to
(1) $n ^ { 2 } + n$
(2) $n$
(3) $\frac { n n + 1 } { 2 }$
(4) $n ^ { 2 }$
jee-main 2023 Q72 View
$\lim_{n \rightarrow \infty} \left\{\left(2^{\frac{1}{2}} - 2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}} - 2^{\frac{1}{5}}\right) \ldots \left(2^{\frac{1}{2}} - 2^{\frac{1}{2n+1}}\right)\right\}$ is equal to
(1) 1
(2) 0
(3) $\sqrt{2}$
(4) $\frac{1}{\sqrt{2}}$
jee-main 2023 Q71 View
The value of $\lim _ { n \rightarrow \infty } \frac { 1 + 2 - 3 + 4 + 5 - 6 + \ldots + ( 3 n - 2 ) + ( 3 n - 1 ) - 3 n } { \sqrt { 2 n ^ { 4 } + 4 n + 3 } - \sqrt { n ^ { 4 } + 5 n + 4 } }$ is
(1) $\frac { \sqrt { 2 } + 1 } { 2 }$
(2) $3 ( \sqrt { 2 } + 1 )$
(3) $\frac { 3 } { 2 } ( \sqrt { 2 } + 1 )$
(4) $\frac { 3 } { 2 \sqrt { 2 } }$
jee-main 2024 Q68 View
$\lim _ { n \rightarrow \infty } \frac { \left( 1 ^ { 2 } - 1 \right) ( n - 1 ) + \left( 2 ^ { 2 } - 2 \right) ( n - 2 ) + \cdots + \left( ( n - 1 ) ^ { 2 } - ( n - 1 ) \right) \cdot 1 } { \left( 1 ^ { 3 } + 2 ^ { 3 } + \cdots \cdots + n ^ { 3 } \right) - \left( 1 ^ { 2 } + 2 ^ { 2 } + \cdots \cdots + n ^ { 2 } \right) }$ is equal to:
(1) $\frac { 2 } { 3 }$
(2) $\frac { 1 } { 3 }$
(3) $\frac { 3 } { 4 }$
(4) $\frac { 1 } { 2 }$
taiwan-gsat 2022 Q8 8 marks View
Suppose two sequences $\langle a_n \rangle$ and $\langle b_n \rangle$ satisfy $b_n + \frac{4n-1}{n} < a_n < 3b_n$ for all positive integers $n$. Given that $\lim_{n \to \infty} a_n = 6$, select the correct options.
(1) $b_n < 6 - \frac{4n-1}{n}$
(2) $b_n > \frac{4n-1}{2n}$
(3) The sequence $\langle b_n \rangle$ may diverge
(4) $a_{10000} < 6.1$
(5) $a_{10000} > 5.9$