For each positive integer n, let yn = (((n+1)(n+2)...(n+n))^(1/n))/n. For x ∈ R, let [x] be the grea

Q: Why can't I just take the limit of each term inside the product individually?

That is a common trap! You cannot distribute a limit across a product of $n$ terms where the number of terms itself is growing to infinity. You must treat the entire product as a single entity, which is why the logarithm is essential to convert the product into a sum that we can handle.

Q: How do I know when to use the Riemann sum method?

Look for the tell-tale signs: a product or sum of $n$ terms, all raised to the power of $1/n$ or multiplied by $1/n$. Whenever you see a sequence that looks like a geometric average of terms, Riemann sums are your best friend. It is the bridge between discrete sequences and continuous calculus.

Q: Why did we get $4/e$ and not something else?

The integral $\int_0^1 \ln(1+x) dx$ evaluates to $2\ln 2 - 1$. Using log properties, this is $\ln 4 - \ln e = \ln(4/e)$. Since we calculated $\ln L$, we must exponentiate to find $L$. The value $4/e$ is approximately $1.47$, and the greatest integer function $[1.47]$ gives us $1$.

Question

For each positive integer $n$, let $y_n = \frac{((n+1)(n+2)...(n+n))^{1/n}}{n}$. For $x \in R$, let $[x]$ be the greatest integer less than or equal to $x$. If $\lim_{n	o\infty} y_n = L$, then the value of $[L]$ is ________.

Sigma Percentile · Accepted Answer

### The Hidden Geometry of Sequences

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of products and powers.

We are looking at the sequence:

$$y_n = \frac{((n+1)(n+2)...(n+n))^{1/n}}{n}$$

It looks intimidating, but in the world of JEE Advanced, intimidation is often just a mask for elegance. Let us peel back that mask.

### Phase 1: The Anatomy of the Sequence

We have a product of $n$ terms, all raised to the power of $1/n$. This is the hallmark of a geometric mean.

When taking the limit as $n 	o \infty$, your mind should immediately jump to the **Riemann sum**. However, Riemann sums are built for addition, not multiplication.

To bridge this gap, we use the natural logarithm. By taking the natural log, we turn the product into a sum because $\ln(a \cdot b) = \ln a + \ln b$. This is the magic wand that transforms our discrete sequence into a continuous integral.

### Phase 2: The Logarithmic Bridge

Let us rewrite our expression by taking the natural log of both sides:

$$\ln(y_n) = \ln \left( \left( \frac{(n+1)(n+2)...(n+n)}{n^n} ight)^{1/n} ight)$$

Using the power rule for logarithms, the $1/n$ factor moves to the front. We now have:

$$\ln(y_n) = \frac{1}{n} \sum_{r=1}^n \ln \left( \frac{n+r}{n} ight)$$

Look closely at the term inside the sum: $\frac{n+r}{n}$ is simply $1 + \frac{r}{n}$. Thus, we arrive at the structure:

$$\ln(y_n) = \frac{1}{n} \sum_{r=1}^n \ln \left( 1 + \frac{r}{n} ight)$$

This is the exact definition of a Riemann sum for the function $f(x) = \ln(1+x)$ over the interval $[0, 1]$.

### Phase 3: The Riemann Transformation

As $n 	o \infty$, this sum transforms into the definite integral:

$$\int_0^1 \ln(1+x) \, dx$$

We are calculating the area under the curve $y = \ln(1+x)$ from $x=0$ to $x=1$. To solve this, we use the substitution $t = 1+x$, which implies $dt = dx$.

When $x=0$, $t=1$; when $x=1$, $t=2$. Our integral becomes:

$$\int_1^2 \ln(t) \, dt$$

The integral of $\ln(t)$ is $t \ln(t) - t$. Evaluating this from $1$ to $2$:

$$(2 \ln 2 - 2) - (1 \ln 1 - 1) = 2 \ln 2 - 2 + 1 = 2 \ln 2 - 1$$

### Phase 4: The Final Reveal

We have found that $\ln(L) = 2 \ln 2 - 1$. Using the properties of logarithms, $2 \ln 2 = \ln(2^2) = \ln 4$, and $1 = \ln e$.

Therefore:

$$\ln(L) = \ln 4 - \ln e = \ln\left(\frac{4}{e}ight)$$

This implies that $L = \frac{4}{e}$. Since $e \approx 2.718$, we find $L \approx \frac{4}{2.718} \approx 1.47$.

The question asks for the greatest integer less than or equal to $L$, denoted as $[L]$. Thus, $[1.47] = 1$.

**The final answer is 1.**

Enter Numerical Value:

Visualized Solution

Analyze $y_{n}$

Rewrite the Denominator

Combine the Terms

Express as a Product

Take Natural Logarithm

Riemann Sum to Integral

Form the Definite Integral

Integration by Substitution

New Integral Limits

Integrate $ln (t)$

Evaluate Limits

Simplify $ln L$

Find $L$

Calculate $[L]$

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

The Hidden Geometry of Sequences

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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

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

Let $[\cdot]$ denote the greatest integer function and $f (x) = lim_{n \to \infty} \frac{1}{n ^{3}} \sum_{k = 1}^{n} [\frac{k ^{2}}{3 ^{x}}]$ . Then $12 \sum_{j = 1}^{\infty} f (j)$ 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:

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

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

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

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

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

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

Enter Numerical Value:

Visualized Solution

Analyze .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } yn​

Rewrite the Denominator

Combine the Terms

Express as a Product

Take Natural Logarithm

Riemann Sum to Integral

Form the Definite Integral

Integration by Substitution

New Integral Limits

Integrate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ln(t)

Evaluate Limits

Simplify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } lnL

Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [L]

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

The Hidden Geometry of Sequences

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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