How do I handle limits like n factorial to the 1 over n divided by n?

Take natural log first. This converts a product to a sum. Divide by n to get a Riemann sum for the integral of log f(x). Evaluate the integral and exponentiate.

Definite Integral as Limit of a Sum

Q: Why is the definite integral defined as a limit of a sum?

Historically it was defined this way to formalise area under a curve. Riemann sums approximate area by rectangles; as the number of rectangles increases, the approximation becomes exact. The Fundamental Theorem later showed antiderivatives evaluate this limit efficiently, but the sum definition is what integral originally meant.

Q: How do I recognise a Riemann-sum limit in a JEE problem?

Look for a limit as n goes to infinity of a sum, each term containing 1/n and a function of r/n. If the summand can be written as (1/n) times f(r/n), convert it to a definite integral.

Q: What are the correct limits of integration?

The lower limit is the limit of r/n at the smallest index. The upper limit is the limit of r/n at the largest index. If r runs from 1 to n, limits are 0 to 1. If r runs to 2n, upper limit is 2.

Q: Does it matter whether the sum index starts at 0 or 1?

No, the limit is the same. Both starting indices converge to the same integral because the difference is a single term whose contribution vanishes in the limit.

Q: What if the sum indices depend on n in more complex ways?

Rewrite so every r appears as r/n and every n combines with r to form a ratio. If the index runs from alpha n to beta n, the integral limits become alpha and beta.

Q: Can this method fail?

Yes, if the summand does not admit the standard form or if f is not Riemann integrable on the target interval. Rare in JEE. If a sum resists conversion, look for telescoping, partial fractions, or generating functions.

Q: Is the Riemann sum definition still used in modern mathematics?

Yes. In numerical analysis, Riemann-like sums such as trapezoidal, Simpson's, and Gaussian quadrature are the backbone of practical integration on computers. In pure mathematics, Lebesgue, Riemann-Stieltjes, and stochastic integrals all build on this idea.

Q: Why does dividing by n turn a sum into an integral?

Because 1/n plays the role of the strip width delta-x when the interval is 0 to 1. The sum measures the total area of n rectangles, and the limit as n goes to infinity shrinks the strip width to zero.

MathsIntegralsFor JEE aspirants

The definite integral as a limit of a sum is the original Riemann definition that Newton-Leibnitz shortcuts. If $f (x)$ is continuous on $[a, b]$ , then $\int_{a}^{b} f (x) d x = lim_{n \to \infty} h \sum_{r = 0}^{n - 1} f (a + r h)$ , where $h = \frac{b - a}{n}$ . The converse is equally powerful: many JEE limits of the form $lim_{n \to \infty} \frac{1}{n} \sum f (\frac{r}{n})$ can be converted directly to a definite integral $\int_{0}^{1} f (x) d x$ and evaluated in closed form. This is a very frequent JEE Main and Advanced pattern.

Key Formulas - Quick Reference

Riemann definition: $\int_{a}^{b} f (x) d x = n \to \infty lim r = 0 \sum n - 1 (\frac{b - a}{n}) f (a + \frac{( b - a ) r}{n})$
Standard form on $[0, 1]$ : $n \to \infty lim \frac{1}{n} r = 0 \sum n - 1 f (\frac{r}{n}) = \int_{0}^{1} f (x) d x$
Alternate index limit (also $[0, 1]$ ): $n \to \infty lim \frac{1}{n} r = 1 \sum n f (\frac{r}{n}) = \int_{0}^{1} f (x) d x$
General limits $[0, p]$ : $n \to \infty lim \frac{1}{n} r = 1 \sum p n f (\frac{r}{n}) = \int_{0}^{p} f (x) d x$
Log-of-product form: $n \to \infty lim (r = 1 \prod n a_{r})^{1/ n}$ often reduces via $ln$ to $\int_{0}^{1} ln f (x) d x$

1. The Riemann Sum Definition

Let $f (x)$ be a continuous real-valued function on the closed interval $[a, b]$ . Divide $[a, b]$ into $n$ equal parts by the points

a, a + h, a + 2 h, \dots, a + (n - 1) h, a + nh = b, where h = \frac{b - a}{n} .

Let $S_{n}$ denote the total area of the $n$ rectangles obtained by taking the function value at the left endpoint of each strip:

S_{n} = h f (a) + h f (a + h) + h f (a + 2 h) + \dots + h f (a + (n - 1) h)

As $n \to \infty$ , the rectangles get thinner and $S_{n}$ approaches the exact area under the curve, giving the Riemann definition:

\int_{a}^{b} f (x) d x = n \to \infty lim S_{n} = n \to \infty lim r = 0 \sum n - 1 h f (a + r h) = n \to \infty lim r = 0 \sum n - 1 (\frac{b - a}{n}) f (a + \frac{( b - a ) r}{n})

Figure 1: Rectangular approximation of

\int_{a}^{b} f (x) d x

. Sum of rectangle areas

\to

exact integral as

n \to \infty

(strip width

h \to 0

Right-endpoint version: You can also write

S_{n} = h [f (a + h) + f (a + 2 h) + \dots + f (a + nh)]

, giving

\int_{a}^{b} f (x) d x = lim_{n \to \infty} \sum_{r = 1}^{n} \frac{b - a}{n} f (a + \frac{( b - a ) r}{n})

. Both indexing choices give the same limit.

2. Converting a Sum to an Integral

The Standard Recipe

To convert a limit-of-sum expression into a definite integral, follow these steps:

Rewrite the sum so it looks like $\sum \frac{1}{n} f (\frac{r}{n})$ . Factor $\frac{1}{n}$ out cleanly.
Replace $\frac{r}{n}$ by $x$ , $\frac{1}{n}$ by $d x$ , and $n \to \infty lim \sum$ by $\int$ .
Determine the lower limit as $n \to \infty lim \frac{r}{n}$ at the smallest value of $r$ , and the upper limit as $n \to \infty lim \frac{r}{n}$ at the largest value of $r$ .
Evaluate the resulting definite integral.

Common Special Cases

$n \to \infty lim \frac{1}{n} r = 1 \sum n f (\frac{r}{n}) = \int_{0}^{1} f (x) d x$ - the workhorse form
$n \to \infty lim \frac{1}{n} r = 1 \sum p n f (\frac{r}{n}) = \int_{0}^{p} f (x) d x$ - when the top index is a multiple of $n$
$n \to \infty lim \frac{1}{n} r = 0 \sum n - 1 f (\frac{r}{n}) = \int_{0}^{1} f (x) d x$ - starting from $r = 0$ gives the same limit

3. Solved Examples

Solved Example 1

Show that

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

Solution:

Rewrite the sum by factoring $1/ n$ from each term:

S = r = 1 \sum n \frac{1}{n + r} = r = 1 \sum n \frac{1}{n ( 1 + \frac{r}{n} )} = \frac{1}{n} r = 1 \sum n \frac{1}{1 + \frac{r}{n}}

Replace $\frac{r}{n} \to x$ , $\frac{1}{n} \to d x$ . Lower limit: $lim \frac{r}{n}$ at $r = 1$ is $0$ . Upper limit: $lim \frac{r}{n}$ at $r = n$ is $1$ .

n \to \infty lim S = \int_{0}^{1} \frac{d x}{1 + x} = [ln (1 + x)]_{0}^{1} = ln 2

Solved Example 2

Evaluate

n \to \infty lim [\frac{1}{1 + n} + \frac{1}{2 + n} + \frac{1}{3 + n} + \dots + \frac{1}{2 n}]

Solution:

This is the same as Example 1 rewritten. Let $S = r = 1 \sum n \frac{1}{r + n} = \frac{1}{n} r = 1 \sum n \frac{1}{\frac{r}{n} + 1}$ .

n \to \infty lim S = \int_{0}^{1} \frac{d x}{x + 1} = ln 2

Solved Example 3

Show that

n \to \infty lim \frac{1 ^{p} + 2 ^{p} + 3 ^{p} + \dots + n ^{p}}{n ^{p + 1}} = \frac{1}{p + 1}

for

p > 0

Solution:

Rewrite:

\frac{1 ^{p} + 2 ^{p} + \dots + n ^{p}}{n ^{p + 1}} = r = 1 \sum n \frac{r ^{p}}{n \cdot n ^{p}} = \frac{1}{n} r = 1 \sum n (\frac{r}{n})^{p}

Take $f (x) = x^{p}$ ; lower limit $0$ , upper limit $1$ :

n \to \infty lim \frac{1}{n} r = 1 \sum n (\frac{r}{n})^{p} = \int_{0}^{1} x^{p} d x = [\frac{x ^{p + 1}}{p + 1}]_{0}^{1} = \frac{1}{p + 1}

Solved Example 4

Evaluate

n \to \infty lim r = 1 \sum 2 n \frac{n + r}{n ^{2} + r ^{2}}

Solution:

Divide numerator and denominator by $n^{2}$ :

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

Lower limit: $lim r / n$ at $r = 1$ is $0$ . Upper limit: $lim r / n$ at $r = 2 n$ is $2$ .

= \int_{0}^{2} \frac{1 + x}{1 + x ^{2}} d x = \int_{0}^{2} \frac{d x}{1 + x ^{2}} + \frac{1}{2} \int_{0}^{2} \frac{2 x d x}{1 + x ^{2}}

= [tan^{- 1} x]_{0}^{2} + \frac{1}{2} [ln (1 + x^{2})]_{0}^{2} = tan^{- 1} 2 + \frac{1}{2} ln 5

Solved Example 5

Evaluate

n \to \infty lim (\frac{n !}{n ^{n}})^{1/ n}

Solution:

Let $y = n \to \infty lim (\frac{n !}{n ^{n}})^{1/ n}$ . Take logarithm:

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

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

Convert to integral with $f (x) = ln x$ , limits $0$ to $1$ :

ln y = \int_{0}^{1} ln x d x = [x ln x - x]_{0}^{1} = (0 - 1) - x \to 0^{+} lim (x ln x) + 0 = - 1 - 0 = - 1

Therefore $y = \frac{1}{e}$ .

Solved Example 6

Evaluate

n \to \infty lim [\frac{1}{n ^{2}} + \frac{1}{n ^{2} + n} + \frac{1}{n ^{2} + 2 n} + \dots + \frac{1}{n ^{2} + n ^{2}}]

Solution:

The general term is $\frac{1}{n ^{2} + r n}$ for $r = 0, 1, \dots, n$ . Factor $n$ from the square root:

r = 0 \sum n \frac{1}{n ^{2} + r n} = r = 0 \sum n \frac{1}{n 1 + \frac{r}{n}} = \frac{1}{n} r = 0 \sum n \frac{1}{1 + \frac{r}{n}}

Lower limit: $r = 0 \Rightarrow 0$ . Upper limit: $r = n \Rightarrow 1$ .

lim = \int_{0}^{1} \frac{d x}{1 + x} = [2 1 + x]_{0}^{1} = 2 (2 - 1)

Solved Example 7

Evaluate

n \to \infty lim \frac{1}{n ^{2}} [sin^{3} \frac{π}{4 n} + 2 sin^{3} \frac{2 π}{4 n} + 3 sin^{3} \frac{3 π}{4 n} + \dots + n sin^{3} \frac{nπ}{4 n}]

Solution:

General term: $\frac{r}{n ^{2}} sin^{3} \frac{r π}{4 n} = \frac{1}{n} \cdot \frac{r}{n} \cdot sin^{3} (\frac{π}{4} \cdot \frac{r}{n})$ .

Let $x = r / n$ ; then

lim = \int_{0}^{1} x sin^{3} (\frac{π x}{4}) d x

Use $sin^{3} u = \frac{3 sin u - sin 3 u}{4}$ . Let $u = π x /4$ , so $d u = (π /4) d x$ , i.e., $d x = (4/ π) d u$ and $x = 4 u / π$ ; limits $u \in [0, π /4]$ :

I = \int_{0}^{π /4} \frac{4 u}{π} \cdot \frac{3 sin u - sin 3 u}{4} \cdot \frac{4}{π} d u = \frac{4}{π ^{2}} \int_{0}^{π /4} u (3 sin u - sin 3 u) d u

Integrate by parts each piece: $\int_{0}^{π /4} u sin u d u = [- u cos u + sin u]_{0}^{π /4} = - \frac{π}{4 2} + \frac{1}{2} = \frac{1}{2} (1 - \frac{π}{4})$ .

And $\int_{0}^{π /4} u sin 3 u d u = [- \frac{u cos 3 u}{3} + \frac{sin 3 u}{9}]_{0}^{π /4} = - \frac{π}{12} cos \frac{3 π}{4} + \frac{1}{9} sin \frac{3 π}{4} = \frac{π}{12 2} + \frac{1}{9 2}$ .

Combining:

I = \frac{4}{π ^{2}} [3 \cdot \frac{1}{2} (1 - \frac{π}{4}) - (\frac{π}{12 2} + \frac{1}{9 2})] = \frac{2}{9 π ^{2}} (52 - 15 π)

Common Mistakes to Avoid

Watch out

Wrong limits of integration. The lower limit is $lim_{n \to \infty} (r / n)$ at the smallest $r$ in the sum (usually 0), and the upper limit is $lim_{n \to \infty} (r / n)$ at the largest $r$ . Missing this step is the most common error.
Forgetting to isolate $\frac{1}{n}$ cleanly. The pattern is $\sum \frac{1}{n} \cdot f (r / n)$ , not $\sum \frac{a}{n} \cdot f (r / n)$ with a stray constant.
Using the sum-to-integral method on a divergent or non-Riemann-sum expression. It requires the sum to be of the exact form $\frac{1}{n} \sum f (r / n)$ up to a substitution.
Skipping the log trick for products. Limits like $(\prod)^{1/ n}$ or $(n!)^{1/ n} / n$ typically need $ln$ to convert to a Riemann sum.
Assuming the sum indices matter. Whether $r$ runs $0$ to $n - 1$ or $1$ to $n$ , the limit is the same. But the upper index affects the upper limit of integration when it is not simply $n$ (e.g., $2 n$ gives upper limit $2$ ).
Trying to compute a sum directly instead of recognising the Riemann pattern. If a JEE problem has $\frac{1}{n}$ and $\frac{r}{n}$ appearing together with a limit, convert to an integral.

Frequently Asked Questions

Q1. Why is the definite integral defined as a limit of a sum?

Historically, the integral was defined this way to formalise the notion of area under a curve. Riemann sums approximate the area by rectangles; as the number of rectangles increases, the approximation gets exact. The Fundamental Theorem later showed that antiderivatives evaluate this limit efficiently, but the sum definition is what "integral" originally meant and is still the theoretical foundation.

Q2. How do I recognise a Riemann-sum limit in a JEE problem?

Look for these signals: (a) a limit as $n \to \infty$ of a sum, (b) each term contains $\frac{1}{n}$ or divides by $n$ , (c) each term contains $\frac{r}{n}$ or a function of $r / n$ , (d) the summand can be written as $\frac{1}{n} f (r / n)$ after algebraic manipulation. If all four are present, convert to $\int f (x) d x$ .

Q3. What are the correct limits of integration?

The lower limit is the limiting value of $r / n$ at the smallest index in your sum. If $r$ starts at 1, that limit is 0. The upper limit is the limiting value of $r / n$ at the largest index. If $r$ goes up to $n$ , the upper limit is 1; if it goes to $2 n$ , the upper limit is 2; if it goes to $p n$ , the upper limit is $p$ .

Q4. Does it matter whether the sum index starts at 0 or 1?

No, the limit is the same. Both $\frac{1}{n} r = 0 \sum n - 1 f (r / n)$ and $\frac{1}{n} r = 1 \sum n f (r / n)$ converge to $\int_{0}^{1} f (x) d x$ . The difference between the two is a single term whose contribution vanishes in the limit.

Q5. How do I handle limits of the form $(n!)^{1/ n} / n$ or a product raised to $1/ n$ ?

Take the natural log first. This converts a product to a sum. Then divide by $n$ so the expression becomes $\frac{1}{n} \sum ln f (r / n)$ , which is a Riemann sum for $\int_{0}^{1} ln f (x) d x$ . Evaluate the integral, then exponentiate the result.

Q6. What if the sum indices depend on $n$ in more complex ways?

Rewrite so every occurrence of $r$ appears as $r / n$ (or a function of it) and every occurrence of $n$ combines with $r$ to form a ratio. If the index runs from $α n$ to $β n$ , the integral limits are $α$ and $β$ . Adjust the algebra until the sum takes the standard $\frac{1}{n} f (r / n)$ shape.

Q7. Can this method fail?

Yes, in two situations: (a) if the summand does not admit the $\frac{1}{n} f (r / n)$ form (some sums genuinely need direct evaluation), and (b) if $f$ is not Riemann integrable on the target interval (this rarely arises in JEE, since JEE integrands are continuous). If a sum resists conversion, look for telescoping, partial fractions, or a generating-function identity.

Q8. Is the Riemann sum definition still used in modern mathematics?

Yes. In numerical analysis, Riemann-like sums (trapezoidal, Simpson's, Gaussian quadrature) are the backbone of practical integration on computers. In pure mathematics, more general integrals (Lebesgue, Riemann-Stieltjes, Ito for stochastic processes) all build on the sum-limit idea. For JEE, Riemann sums are important both as a definition and as a tool to convert limits to integrals.

Q9. Why does dividing by $n$ turn a sum into an integral?

Because $\frac{1}{n}$ plays the role of the strip width $Δ x = \frac{b - a}{n}$ when the interval is $[0, 1]$ ( $b - a = 1$ ). The sum then measures the total area of $n$ rectangles, and the limit $n \to \infty$ makes the strip width shrink to zero, converting the sum into an integral by the Riemann definition.

Previous year questions on Definite Integral as Limit of a Sum

5 questions from past papers, each with a step-by-step solution.

Ready to master Integrals?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test

1. The Riemann Sum Definition

2. Converting a Sum to an Integral

The Standard Recipe

Common Special Cases

3. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

Q1. Why is the definite integral defined as a limit of a sum?

Q2. How do I recognise a Riemann-sum limit in a JEE problem?

Q3. What are the correct limits of integration?

Q4. Does it matter whether the sum index starts at 0 or 1?

Q5. How do I handle limits of the form (n!)1/n/n or a product raised to 1/n?

Q6. What if the sum indices depend on n in more complex ways?

Q7. Can this method fail?

Q8. Is the Riemann sum definition still used in modern mathematics?

Q9. Why does dividing by n turn a sum into an integral?

Previous year questions on Definite Integral as Limit of a Sum

Ready to master Integrals?

Q5. How do I handle limits of the form $(n!)^{1/ n} / n$ or a product raised to $1/ n$ ?

Q6. What if the sum indices depend on $n$ in more complex ways?

Q9. Why does dividing by $n$ turn a sum into an integral?