Definite Integral as a Limit of a Sum

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JEE Main 2016

LEVELJEE Main

$lim_{n \to \infty} (\frac{( n + 1 ) ( n + 2 ) \dots 3 n}{n ^{2 n}})^{\frac{1}{n}}$ is equal to

$lim_{n \to \infty} [\frac{1}{n ^{2}} sec^{2} \frac{1}{n ^{2}} + \frac{2}{n ^{2}} sec^{2} \frac{4}{n ^{2}} + \dots + \frac{1}{n} sec^{2} 1]$ equals

(A)

\frac{1}{2} sec 1

(B)

\frac{1}{2} csc 1

(C)

tan 1

(D)

\frac{1}{2} tan 1

JEE Main 2004

LEVELJEE Main

$lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n} e^{r / n}$ is

(A)

e + 1

(B)

e - 1

(C)

1 - e

(D)

e

JEE Main 2003

LEVELJEE Main

$lim_{n \to \infty} \frac{1 ^{4} + 2 ^{4} + 3 ^{4} + \dots + n ^{4}}{n ^{5}} - lim_{n \to \infty} \frac{1 ^{3} + 2 ^{3} + 3 ^{3} + \dots + n ^{3}}{n ^{5}}$

$lim_{n \to \infty} \frac{1 ^{p} + 2 ^{p} + 3 ^{p} + \dots + n ^{p}}{n ^{p + 1}}$

(A)

\frac{1}{p + 1}

(B)

\frac{1}{1 - p}

(C)

\frac{1}{p - 1} - \frac{1}{p}

(D)

\frac{1}{p + 2}

JEE Advanced 2016

LEVELJEE Advanced

Let $f (x) = lim_{n \to \infty} (\frac{n ^{n} ( x + n ) ( x + \frac{n}{2} ) ... ( x + \frac{n}{n} )}{n ! ( x ^{2} + n ^{2} ) ( x ^{2} + \frac{n ^{2}}{4} ) ... ( x ^{2} + \frac{n ^{2}}{n ^{2}} )})^{x / n}$ , for all $x > 0$ . Then

* Multiple Correct Options

(A)

f (\frac{1}{2}) \geq f (1)

(B)

f (\frac{1}{3}) \leq f (\frac{2}{3})

(C)

f^{'} (2) \leq 0

(D)

\frac{f ^{'} ( 3 )}{f ( 3 )} \geq \frac{f ^{'} ( 2 )}{f ( 2 )}

JEE Advanced 2013

LEVELJEE Main

For $a \in R$ (the set of all real numbers), $a \neq = - 1, lim_{n \to \infty} \frac{( 1 ^{a} + 2 ^{a} + \dots + n ^{a} )}{( n + 1 ) ^{a - 1} [( na + 1 ) + ( na + 2 ) + \dots + ( na + n )]} = \frac{1}{60}$ . Then $a = .........$

* Multiple Correct Options

(A)

(B)

(C)

- 15/2

(D)

- 17/2

JEE Advanced 2008

LEVELJEE Advanced

Let $S_{n} = \sum_{k = 1}^{n} \frac{n}{n ^{2} + k n + k ^{2}}$ and $T_{n} = \sum_{k = 0}^{n - 1} \frac{n}{n ^{2} + k n + k ^{2}}$ for $n = 1, 2, 3, \dots$ . Then,

* Multiple Correct Options

(A)

S_{n} < \frac{π}{3 3}

(B)

S_{n} > \frac{π}{3 3}

(C)

T_{n} < \frac{π}{3 3}

(D)

T_{n} > \frac{π}{3 3}

JEE Advanced 1997

LEVELJEE Main

$lim_{n \to \infty} \sum_{r = 1}^{2 n} \frac{r}{n ^{2} + r ^{2}}$ equals

(A)

1 + 5

(B)

- 1 + 5

(C)

- 1 + 2

(D)

1 + 2

JEE Advanced 1981

LEVELJEE Main

Show that: $lim_{n \to \infty} (\frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{6 n}) = lo g 6$ .

JEE Main 2019 (12 January)

LEVELJEE Main

$lim_{n \to \infty} (\frac{n}{n ^{2} + 1 ^{2}} + \frac{n}{n ^{2} + 2 ^{2}} + \dots + \frac{1}{5 n})$ is equal to :

(A)

\frac{π}{4}

(B)

tan^{- 1} (2)

(C)

tan^{- 1} (3)

(D)

\frac{π}{2}

JEE Main 2019 (10 April Shift 1)

LEVELJEE Main

$lim_{n \to \infty} (\frac{( n + 1 ) ^{1/3}}{n ^{4/3}} + \frac{( n + 2 ) ^{1/3}}{n ^{4/3}} + \dots + \frac{( 2 n ) ^{1/3}}{n ^{4/3}})$ is equal to :

(A)

\frac{4}{3} (2)^{4/3}

(B)

\frac{3}{4} (2)^{4/3} - \frac{3}{4}

(C)

\frac{3}{4} (2)^{3/4} - \frac{3}{4}

(D)

\frac{3}{4} (2)^{4/3}

JEE Main 2021 (27 Aug Shift 1)

LEVELJEE Advanced

If $U_{n} = (1 + \frac{1}{n ^{2}}) (1 + \frac{2 ^{2}}{n ^{2}})^{2} \dots (1 + \frac{n ^{2}}{n ^{2}})^{n}$ , then $lim_{n \to \infty} (U_{n})^{- 4/ n^{2}}$ is equal to :

(A)

\frac{e ^{2}}{16}

(B)

\frac{4}{e}

(C)

\frac{16}{e ^{2}}

(D)

\frac{4}{e ^{2}}

JEE Main 2021 (26 Aug Shift 1)

LEVELJEE Main

The value of $lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{n ^{2}}{n ^{2} + 4 r ^{2}}$ is:

(A)

\frac{1}{2} tan^{- 1} (2)

(B)

\frac{1}{2} tan^{- 1} (4)

(C)

tan^{- 1} (4)

(D)

\frac{1}{4} tan^{- 1} (4)

JEE Main 2021 (27 July Shift 1)

LEVELJEE Main

The value of $lim_{n \to \infty} \frac{1}{n} \sum_{j = 1}^{n} \frac{( 2 j - 1 ) + 8 n}{( 2 j - 1 ) + 4 n}$ is equal to :

(A)

5 + lo g_{e} (\frac{3}{2})

(B)

2 - lo g_{e} (\frac{2}{3})

(C)

3 + 2 lo g_{e} (\frac{2}{3})

(D)

1 + 2 lo g_{e} (\frac{3}{2})

JEE Main 2021 (20 July Shift 2)

LEVELJEE Main

If $f : R \to R$ is given by $f (x) = x + 1$ , then the value of $lim_{n \to \infty} \frac{1}{n} [f (0) + f (\frac{5}{n}) + f (\frac{10}{n}) + \dots + f (\frac{5 ( n - 1 )}{n})]$ is:

(A)

\frac{3}{2}

(B)

\frac{5}{2}

(C)

\frac{1}{2}

(D)

\frac{7}{2}

JEE Main 2021 (16 March Shift 1)

LEVELJEE Advanced

Let $f : (0, 2) \to R$ be defined as $f (x) = lo g_{2} (1 + tan (\frac{π x}{4}))$ . Then, $lim_{n \to \infty} \frac{2}{n} (f (\frac{1}{n}) + f (\frac{2}{n}) + \dots + f (1))$ is equal to ______

JEE Main 2022 (24 June Shift 2)

LEVELJEE Advanced

$lim_{n \to \infty} (\frac{n ^{2}}{( n ^{2} + 1 ) ( n + 1 )} + \frac{n ^{2}}{( n ^{2} + 4 ) ( n + 2 )} + \frac{n ^{2}}{( n ^{2} + 9 ) ( n + 3 )} + \dots + \frac{n ^{2}}{( n ^{2} + n ^{2} ) ( n + n )})$ is equal to

(A)

\frac{π}{8} + \frac{1}{4} lo g_{e} 2

(B)

\frac{π}{4} + \frac{1}{8} lo g_{e} 2

(C)

\frac{π}{4} - \frac{1}{8} lo g_{e} 2

(D)

\frac{π}{8} + lo g_{e} 2

JEE Main 2022 (26 July Shift 1)

LEVELJEE Main

If $a = lim_{n \to \infty} \sum_{k = 1}^{n} \frac{2 n}{n ^{2} + k ^{2}}$ and $f (x) = \frac{1 - c o s x}{1 + c o s x}, x \in (0, 1)$ , then :

(A)

22 f (\frac{a}{2}) = f^{'} (\frac{a}{2})

(B)

f (\frac{a}{2}) f^{'} (\frac{a}{2}) = 2

(C)

2 f (\frac{a}{2}) = f^{'} (\frac{a}{2})

(D)

f (\frac{a}{2}) = 2 f^{'} (\frac{a}{2})

JEE Main 2022 (25 July Shift 1)

LEVELJEE Main

If $lim_{n \to \infty} \frac{( n + 1 ) ^{k - 1}}{n ^{k + 1}} [(nk + 1) + (nk + 2) + \dots + (nk + n)] = 33 \cdot lim_{n \to \infty} \frac{1}{n ^{k + 1}} [1^{k} + 2^{k} + 3^{k} + \dots + n^{k}]$ , then the integral value of $k$ is equal to ______.

JEE Main 2023 (30 January Shift 2)

LEVELJEE Main

$lim_{n \to \infty} \frac{3}{n} \sum_{r = 0}^{n - 1} (2 + \frac{r}{n})^{2}$ is equal to

(A)

12

(B)

\frac{19}{3}

(C)

0

(D)

19

JEE Main 2023 (01 February Shift 1)

LEVELJEE Main

$lim_{n \to \infty} (\frac{1}{1 + n} + \frac{1}{2 + n} + \frac{1}{3 + n} + ... + \frac{1}{2 n})$ is equal to :-

(A)

0

(B)

lo g_{e} 2

(C)

lo g_{e} (3/2)

(D)

lo g_{e} (2/3)

JEE Main 2023 (13 Apr Shift 1)

LEVELJEE Main

Among $(S 1) : lim_{n \to \infty} \frac{1}{n ^{2}} (2 + 4 + 6 + \dots + 2 n) = 1$ and $(S 2) : lim_{n \to \infty} \frac{1}{n ^{16}} (1^{15} + 2^{15} + 3^{15} + \dots + n^{15}) = \frac{1}{16}$

(A)

Both (S1) and (S2) are true

(B)

Only (S1) is true

(C)

Both (S1) and (S2) are false

(D)

Only (S2) is true

JEE Main 2024 (09 April Shift 1)

LEVELJEE Advanced

Let $lim_{n \to \infty} (\frac{n}{n ^{4} + 1} - \frac{2 n}{( n ^{2} + 1 ) n ^{4} + 1} + \frac{n}{n ^{4} + 16} - \frac{8 n}{( n ^{2} + 4 ) n ^{4} + 16} + \dots + \frac{n}{n ^{4} + n ^{4}} - \frac{2 n \cdot n ^{2}}{( n ^{2} + n ^{2} ) n ^{4} + n ^{4}})$ be $\frac{π}{k}$ , using only the principal values of the inverse trigonometric functions. Then $k^{2}$ is equal to ________

JEE Main 2024 (30 Jan Shift 1)

LEVELJEE Main

The value of $lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n ^{3}}{( n ^{2} + k ^{2} ) ( n ^{2} + 3 k ^{2} )}$ is:

(A)

\frac{( 2 3 + 3 ) π}{24}

(B)

\frac{13 π}{8 ( 4 3 + 3 )}

(C)

\frac{13 ( 2 3 - 3 ) π}{8}

(D)

\frac{π}{8 ( 2 3 + 3 )}

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$lim_{n \to \infty} (\frac{( n + 1 ) ( n + 2 ) \dots 3 n}{n ^{2 n}})^{\frac{1}{n}}$ is equal to

$lim_{n \to \infty} [\frac{1}{n ^{2}} sec^{2} \frac{1}{n ^{2}} + \frac{2}{n ^{2}} sec^{2} \frac{4}{n ^{2}} + \dots + \frac{1}{n} sec^{2} 1]$ equals

$lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n} e^{r / n}$ is

$lim_{n \to \infty} \frac{1 ^{4} + 2 ^{4} + 3 ^{4} + \dots + n ^{4}}{n ^{5}} - lim_{n \to \infty} \frac{1 ^{3} + 2 ^{3} + 3 ^{3} + \dots + n ^{3}}{n ^{5}}$

$lim_{n \to \infty} \frac{1 ^{p} + 2 ^{p} + 3 ^{p} + \dots + n ^{p}}{n ^{p + 1}}$

Let $f (x) = lim_{n \to \infty} (\frac{n ^{n} ( x + n ) ( x + \frac{n}{2} ) ... ( x + \frac{n}{n} )}{n ! ( x ^{2} + n ^{2} ) ( x ^{2} + \frac{n ^{2}}{4} ) ... ( x ^{2} + \frac{n ^{2}}{n ^{2}} )})^{x / n}$ , for all $x > 0$ . Then

For $a \in R$ (the set of all real numbers), $a \neq = - 1, lim_{n \to \infty} \frac{( 1 ^{a} + 2 ^{a} + \dots + n ^{a} )}{( n + 1 ) ^{a - 1} [( na + 1 ) + ( na + 2 ) + \dots + ( na + n )]} = \frac{1}{60}$ . Then $a = .........$

Let $S_{n} = \sum_{k = 1}^{n} \frac{n}{n ^{2} + k n + k ^{2}}$ and $T_{n} = \sum_{k = 0}^{n - 1} \frac{n}{n ^{2} + k n + k ^{2}}$ for $n = 1, 2, 3, \dots$ . Then,

$lim_{n \to \infty} \sum_{r = 1}^{2 n} \frac{r}{n ^{2} + r ^{2}}$ equals

Show that: $lim_{n \to \infty} (\frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{6 n}) = lo g 6$ .

$lim_{n \to \infty} (\frac{n}{n ^{2} + 1 ^{2}} + \frac{n}{n ^{2} + 2 ^{2}} + \dots + \frac{1}{5 n})$ is equal to :

$lim_{n \to \infty} (\frac{( n + 1 ) ^{1/3}}{n ^{4/3}} + \frac{( n + 2 ) ^{1/3}}{n ^{4/3}} + \dots + \frac{( 2 n ) ^{1/3}}{n ^{4/3}})$ is equal to :

If $U_{n} = (1 + \frac{1}{n ^{2}}) (1 + \frac{2 ^{2}}{n ^{2}})^{2} \dots (1 + \frac{n ^{2}}{n ^{2}})^{n}$ , then $lim_{n \to \infty} (U_{n})^{- 4/ n^{2}}$ is equal to :

The value of $lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{n ^{2}}{n ^{2} + 4 r ^{2}}$ is:

The value of $lim_{n \to \infty} \frac{1}{n} \sum_{j = 1}^{n} \frac{( 2 j - 1 ) + 8 n}{( 2 j - 1 ) + 4 n}$ is equal to :

If $f : R \to R$ is given by $f (x) = x + 1$ , then the value of $lim_{n \to \infty} \frac{1}{n} [f (0) + f (\frac{5}{n}) + f (\frac{10}{n}) + \dots + f (\frac{5 ( n - 1 )}{n})]$ is:

Let $f : (0, 2) \to R$ be defined as $f (x) = lo g_{2} (1 + tan (\frac{π x}{4}))$ . Then, $lim_{n \to \infty} \frac{2}{n} (f (\frac{1}{n}) + f (\frac{2}{n}) + \dots + f (1))$ is equal to ______

$lim_{n \to \infty} (\frac{n ^{2}}{( n ^{2} + 1 ) ( n + 1 )} + \frac{n ^{2}}{( n ^{2} + 4 ) ( n + 2 )} + \frac{n ^{2}}{( n ^{2} + 9 ) ( n + 3 )} + \dots + \frac{n ^{2}}{( n ^{2} + n ^{2} ) ( n + n )})$ is equal to

If $a = lim_{n \to \infty} \sum_{k = 1}^{n} \frac{2 n}{n ^{2} + k ^{2}}$ and $f (x) = \frac{1 - c o s x}{1 + c o s x}, x \in (0, 1)$ , then :

If $lim_{n \to \infty} \frac{( n + 1 ) ^{k - 1}}{n ^{k + 1}} [(nk + 1) + (nk + 2) + \dots + (nk + n)] = 33 \cdot lim_{n \to \infty} \frac{1}{n ^{k + 1}} [1^{k} + 2^{k} + 3^{k} + \dots + n^{k}]$ , then the integral value of $k$ is equal to ______.

$lim_{n \to \infty} \frac{3}{n} \sum_{r = 0}^{n - 1} (2 + \frac{r}{n})^{2}$ is equal to

$lim_{n \to \infty} (\frac{1}{1 + n} + \frac{1}{2 + n} + \frac{1}{3 + n} + ... + \frac{1}{2 n})$ is equal to :-

Among $(S 1) : lim_{n \to \infty} \frac{1}{n ^{2}} (2 + 4 + 6 + \dots + 2 n) = 1$ and $(S 2) : lim_{n \to \infty} \frac{1}{n ^{16}} (1^{15} + 2^{15} + 3^{15} + \dots + n^{15}) = \frac{1}{16}$

The value of $lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n ^{3}}{( n ^{2} + k ^{2} ) ( n ^{2} + 3 k ^{2} )}$ is: