$\lim_{n \to \infty} \left( \frac{(n+1)(n+2)\dots 3n}{n^{2n}} \right)^{\frac{1}{n}}$ is equal to Options: (A) 9/e^2 (B) 3 \log 3 - 2 (C) 18/e^4 (D) 27/e^2

### Analyzing the Setup We are tasked with evaluating the limit: $$y = \lim_{n \to \infty} \left( \frac{(n+1)(n+2)\dots(3n)}{n^{2n}} \right)^{1/n}$$ Whenever a variable appears in the exponent, the natural logarithm is our most powerful tool. Taking the natural logarithm of both sides, we obtain: $$\ln y = \lim_{n \to \infty} \frac{1}{n} \ln \left( \frac{(n+1)(n+2)\dots(3n)}{n^{2n}} \right)$$ ### The Anatomy of the Product The expression inside the logarithm contains $2n$ terms in the numerator and $n^{2n}$ in the denominator. We can distribute the $n^{2n}$ by assigning one factor of $n$ to each term in the numerator: $$\frac{(n+1)}{n} \cdot \frac{(n+2)}{n} \cdot \dots \cdot \frac{(n+2n)}{n} = \prod_{r=1}^{2n} \left( 1 + \frac{r}{n} \right)$$ Applying the properties of logarithms, the product transforms into a summation: $$\ln y = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{2n} \ln \left( 1 + \frac{r}{n} \right)$$ ### The Riemann Sum Revelation This expression is the classic definition of a **Riemann Sum**. We are calculating the area under the curve $f(x) = \ln(1+x)$ with a width of $dx = 1/n$. As $n \to \infty$, the sum converts into a definite integral. Since the index $r$ ranges from $1$ to $2n$, the limits of integration for $x = r/n$ are from $0$ to $2$: $$\ln y = \int_{0}^{2} \ln(1+x) \, dx$$ ### The Final Resolution To evaluate the integral, we use the substitution $u = 1+x$, which implies $du = dx$. The limits change from $[0, 2]$ to $[1, 3]$: $$\ln y = \int_{1}^{3} \ln(u) \, du$$ The antiderivative of $\ln u$ is $u \ln u - u$. Evaluating this at the boundaries: $$\ln y = [u \ln u - u]_{1}^{3} = (3 \ln 3 - 3) - (1 \ln 1 - 1)$$ Since $\ln 1 = 0$, the expression simplifies to: $$\ln y = 3 \ln 3 - 3 + 1 = 3 \ln 3 - 2$$ To find $y$, we exponentiate both sides: $$y = e^{3 \ln 3 - 2} = e^{\ln(3^3)} \cdot e^{-2} = \frac{27}{e^2}$$ The final result is: $$\mathbf{y = \frac{27}{e^2}}$$

JEE Main 2016

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: $lim_{n \to \infty} (\frac{( n + 1 ) ( n + 2 ) \dots 3 n}{n ^{2 n}})^{\frac{1}{n}}$ is equal to

Q: How did you get the limits of integration from 0 to 2? I always get confused there.

Think of the Riemann Sum definition: $\lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{kn} f(\frac{r}{n}) = \int_0^k f(x) dx$. Here, our sum goes up to $2n$, so $k=2$. The lower limit comes from $r=1$ (which is $1/n \to 0$) and the upper limit comes from $r=2n$ (which is $2n/n = 2$).

Q: Why is the number of terms in the numerator $2n$?

Count them carefully! The numerator is $(n+1), (n+2), \dots, (n+2n)$. This is an arithmetic progression where the number of terms is $(Last - First) + 1$. So, $(n+2n) - (n+1) + 1 = 2n$. Always verify your count before proceeding!

Select Answer:

Visualized Solution

Define the Limit Expression

Let $y = lim_{n \to \infty} (\frac{( n + 1 ) ( n + 2 ) \dots 3 n}{n ^{2 n}})^{\frac{1}{n}}$
We need to evaluate this complex limit involving a product and a variable exponent.

The Logarithmic Approach

The presence of the exponent $\frac{1}{n}$ and a product suggests taking the natural logarithm.
$ln y = ln [lim_{n \to \infty} (\frac{( n + 1 ) ( n + 2 ) \dots 3 n}{n ^{2 n}})^{\frac{1}{n}}]$

Applying Log Properties

Bring the exponent down using the power rule:
$ln y = lim_{n \to \infty} \frac{1}{n} ln (\frac{( n + 1 ) ( n + 2 ) \dots 3 n}{n ^{2 n}})$

Analyzing the Terms

Count the terms in the numerator: from $(n + 1)$ to $3 n$ .
$3 n$ can be written as $(n + 2 n)$ .
Total number of terms in the numerator is $2 n$ .

Distributing the Denominator

The denominator is $n^{2 n}$ , which is exactly $n$ multiplied by itself $2 n$ times.
Pair each of the $2 n$ terms in the numerator with one $n$ from the denominator.
$\frac{n + 1}{n} \cdot \frac{n + 2}{n} \dots \frac{n + 2 n}{n}$

Logarithm of a Product

$ln y = lim_{n \to \infty} \frac{1}{n} ln (\frac{n + 1}{n} \cdot \frac{n + 2}{n} \dots \frac{n + 2 n}{n})$
Using $ln (a \cdot b) = ln a + ln b$ , convert the product to a sum.

Sigma Notation

Express the sum compactly using Sigma notation:
$ln y = lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{2 n} ln (\frac{n + r}{n})$
Simplify the fraction: $\frac{n + r}{n} = 1 + \frac{r}{n}$

Riemann Sum Recognition

The expression is now $lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{2 n} ln (1 + \frac{r}{n})$
This perfectly matches the structure of a Riemann Sum: $lim_{n \to \infty} \frac{1}{n} \sum f (\frac{r}{n})$

Converting to Definite Integral

To convert to an integral, we make standard substitutions:
Replace $\frac{r}{n}$ with $x$ .
Replace $\frac{1}{n}$ with $d x$ .
The summation $\sum$ becomes the integral sign $\int$ .

Finding the Limits of Integration

Lower limit: When $r = 1$ , $x = lim_{n \to \infty} \frac{1}{n} = 0$
Upper limit: When $r = 2 n$ , $x = lim_{n \to \infty} \frac{2 n}{n} = 2$

The Definite Integral

The limit expression is now completely transformed:
$ln y = \int_{0}^{2} ln (1 + x) d x$
We need to find the area under the curve $y = ln (1 + x)$ from $x = 0$ to $x = 2$ .

Evaluating the Integral

Let $1 + x = t$ , then $d x = d t$ .
When $x = 0, t = 1$ . When $x = 2, t = 3$ .
$ln y = \int_{1}^{3} ln t d t$

Integration by Parts

The integral of $ln t$ is a standard result derived using integration by parts:
$\int ln t d t = t ln t - t$
Apply the limits: $ln y = [t ln t - t]_{1}^{3}$

Applying the Limits

Substitute upper limit (3): $3 ln 3 - 3$
Substitute lower limit (1): $1 ln 1 - 1 = 0 - 1 = - 1$
$ln y = (3 ln 3 - 3) - (- 1) = 3 ln 3 - 2$

Final Answer

We have $ln y = 3 ln 3 - 2$
Use log properties: $3 ln 3 = ln (3^{3}) = ln 27$
And $2 = 2 ln e = ln (e^{2})$
$ln y = ln 27 - ln (e^{2}) = ln (\frac{27}{e ^{2}})$
Therefore, $y = \frac{27}{e ^{2}}$

00:00 / 00:00

The Sigma Insight: Definite Integral as a Limit of a Sum

Solution Diagram

Analyzing the Setup

We are tasked with evaluating the limit:

y = n \to \infty lim (\frac{( n + 1 ) ( n + 2 ) \dots ( 3 n )}{n ^{2 n}})^{1/ n}

Whenever a variable appears in the exponent, the natural logarithm is our most powerful tool. Taking the natural logarithm of both sides, we obtain:

ln y = n \to \infty lim \frac{1}{n} ln (\frac{( n + 1 ) ( n + 2 ) \dots ( 3 n )}{n ^{2 n}})

The Anatomy of the Product

The expression inside the logarithm contains

2 n

terms in the numerator and

n^{2 n}

in the denominator. We can distribute the

n^{2 n}

by assigning one factor of

n

to each term in the numerator:

\frac{( n + 1 )}{n} \cdot \frac{( n + 2 )}{n} \cdot \dots \cdot \frac{( n + 2 n )}{n} = r = 1 \prod 2 n (1 + \frac{r}{n})

Applying the properties of logarithms, the product transforms into a summation:

ln y = n \to \infty lim \frac{1}{n} r = 1 \sum 2 n ln (1 + \frac{r}{n})

The Riemann Sum Revelation

This expression is the classic definition of a Riemann Sum. We are calculating the area under the curve

f (x) = ln (1 + x)

with a width of

d x = 1/ n

n \to \infty

, the sum converts into a definite integral. Since the index

r

ranges from

1

2 n

, the limits of integration for

x = r / n

are from

0

2

ln y = \int_{0}^{2} ln (1 + x) d x

The Final Resolution

To evaluate the integral, we use the substitution

u = 1 + x

, which implies

d u = d x

. The limits change from

[0, 2]

[1, 3]

ln y = \int_{1}^{3} ln (u) d u

The antiderivative of

ln u

u ln u - u

. Evaluating this at the boundaries:

ln y = [u ln u - u]_{1}^{3} = (3 ln 3 - 3) - (1 ln 1 - 1)

Since

ln 1 = 0

, the expression simplifies to:

ln y = 3 ln 3 - 3 + 1 = 3 ln 3 - 2

To find

y

, we exponentiate both sides:

y = e^{3 l n 3 - 2} = e^{l n (3^{3})} \cdot e^{- 2} = \frac{27}{e ^{2}}

The final result is:

y = \frac{27}{e ^{2}}

Common Traps & FAQs

Wait, why can we just move the limit inside the logarithm? Is that always legal?

Great question! Yes, because the natural logarithm function is continuous. For any continuous function

f

lim_{n \to \infty} f (a_{n}) = f (lim_{n \to \infty} a_{n})

. Since

ln (x)

is continuous for

x > 0

, we can safely slide the limit inside.

How did you get the limits of integration from 0 to 2? I always get confused there.

Think of the Riemann Sum definition:

lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{k n} f (\frac{r}{n}) = \int_{0}^{k} f (x) d x

. Here, our sum goes up to

2 n

, so

k = 2

. The lower limit comes from

r = 1

(which is

1/ n \to 0

) and the upper limit comes from

r = 2 n

(which is

2 n / n = 2

Why is the number of terms in the numerator

2 n

Count them carefully! The numerator is

(n + 1), (n + 2), \dots, (n + 2 n)

. This is an arithmetic progression where the number of terms is

(L a s t - F i r s t) + 1

. So,

(n + 2 n) - (n + 1) + 1 = 2 n

. Always verify your count before proceeding!

Animated Solution for Mathematics - Definite Integration: $lim_{n \to \infty} (\frac{( n + 1 ) ( n + 2 ) \dots 3 n}{n ^{2 n}})^{\frac{1}{n}}$ is equal to

Select Answer:

Visualized Solution

Define the Limit Expression

The Logarithmic Approach

Applying Log Properties

Analyzing the Terms

Distributing the Denominator

Logarithm of a Product

Sigma Notation

Riemann Sum Recognition

Converting to Definite Integral

Finding the Limits of Integration

The Definite Integral

Evaluating the Integral

Integration by Parts

Applying the Limits

Final Answer

The Sigma Insight: Definite Integral as a Limit of a Sum

Analyzing the Setup

The Anatomy of the Product

The Riemann Sum Revelation

The Final Resolution

Similar Questions

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

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

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

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

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

For positive integer $n$ , define $f (n) = n + \frac{16 + 5 n - 3 n ^{2}}{4 n + 3 n ^{2}} + \frac{32 + n - 3 n ^{2}}{8 n + 3 n ^{2}} + \frac{48 - 3 n - 3 n ^{2}}{12 n + 3 n ^{2}} + \dots + \frac{25 n - 7 n ^{2}}{7 n ^{2}}$ . Then, the value of $lim_{n \to \infty} f (n)$ is equal to

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 :

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

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

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

Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limn→∞​(n2n(n+1)(n+2)…3n​)n1​ is equal to

Select Answer:

Visualized Solution

Define the Limit Expression

The Logarithmic Approach

Applying Log Properties

Analyzing the Terms

Distributing the Denominator

Logarithm of a Product

Sigma Notation

Riemann Sum Recognition

Converting to Definite Integral

Finding the Limits of Integration

The Definite Integral

Evaluating the Integral

Integration by Parts

Applying the Limits

Final Answer

The Sigma Insight: Definite Integral as a Limit of a Sum

Analyzing the Setup

The Anatomy of the Product

The Riemann Sum Revelation

The Final Resolution

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limn→∞​((n2+1)(n+1)n2​+(n2+4)(n+2)n2​+(n2+9)(n+3)n2​+⋯+(n2+n2)(n+n)n2​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limn→∞​(n4/3(n+1)1/3​+n4/3(n+2)1/3​+⋯+n4/3(2n)1/3​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limn→∞​(1+n1​+2+n1​+3+n1​+...+2n1​) is equal to :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limn→∞​(n2+12n​+n2+22n​+⋯+5n1​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limn→∞​[n21​sec2n21​+n22​sec2n24​+⋯+n1​sec21] equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For positive integer n, define f(n)=n+4n+3n216+5n−3n2​+8n+3n232+n−3n2​+12n+3n248−3n−3n2​+⋯+7n225n−7n2​. Then, the value of limn→∞​f(n) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of limn→∞​n1​∑j=1n​(2j−1)+4n(2j−1)+8n​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of limn→∞​∑k=1n​(n2+k2)(n2+3k2)n3​ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limn→∞​n514+24+34+⋯+n4​−limn→∞​n513+23+33+⋯+n3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limn→∞​np+11p+2p+3p+⋯+np​

Animated Solution for Mathematics - Definite Integration: $lim_{n \to \infty} (\frac{( n + 1 ) ( n + 2 ) \dots 3 n}{n ^{2 n}})^{\frac{1}{n}}$ is equal to

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

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

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

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

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

For positive integer $n$ , define $f (n) = n + \frac{16 + 5 n - 3 n ^{2}}{4 n + 3 n ^{2}} + \frac{32 + n - 3 n ^{2}}{8 n + 3 n ^{2}} + \frac{48 - 3 n - 3 n ^{2}}{12 n + 3 n ^{2}} + \dots + \frac{25 n - 7 n ^{2}}{7 n ^{2}}$ . Then, the value of $lim_{n \to \infty} f (n)$ is equal to

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 :

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

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

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