Limit as Sum Technique: Converting Summations to Definite Integrals

Introduction

The "Limit as a Sum" technique transforms a limit of a summation into a definite integral using the Riemann sum definition. This powerful method frequently appears in JEE Main and Advanced, allowing you to evaluate complicated limits by converting them into simpler integrals.

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1. Fundamental Principle

Riemann Sum Definition

For a continuous function $f$ on $[a, b]$: $\int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{r=1}^{n} f(x_r) \Delta x,$ where $\Delta x = \frac{b-a}{n}$ and $x_r = a + r\Delta x$.

Key Conversion Formulas

For $[0, 1]$: $\boxed{\lim_{n\to\infty} \frac{1}{n} \sum_{r=1}^{n} f\!\left(\frac{r}{n}\right) = \int_0^1 f(x)\,dx}$

For $[0, k]$: $\lim_{n\to\infty} \frac{k}{n} \sum_{r=1}^{n} f\!\left(\frac{kr}{n}\right) = \int_0^k f(x)\,dx$

General $[a, b]$: $\lim_{n\to\infty} \frac{b-a}{n} \sum_{r=1}^{n} f\!\left(a + \frac{r(b-a)}{n}\right) = \int_a^b f(x)\,dx$

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2. Step-by-Step Method

Step 1: Express the given sum in the form $\sum \frac{1}{n} \, g\!\left(\frac{r}{n}\right)$.
Step 2: Identify $f(x)$ by replacing $\frac{r}{n}$ with $x$.
Step 3: Determine the limits of integration from the range of $\frac{r}{n}$.
Step 4: Write the definite integral and evaluate it.

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3. Worked Examples

Example 1 (Classic Harmonic Type)

Evaluate $\displaystyle \lim_{n\to\infty} \left( \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} \right)$.

Solution:
Write as: $S = \lim_{n\to\infty} \sum_{r=1}^{n} \frac{1}{n+r} = \lim_{n\to\infty} \sum_{r=1}^{n} \frac{1}{n} \cdot \frac{1}{1 + \frac{r}{n}}.$ Here $f(x) = \frac{1}{1+x}$, and $\frac{r}{n}$ ranges from 0 to 1.
Thus: $S = \int_0^1 \frac{dx}{1+x} = \big[\ln(1+x)\big]_0^1 = \ln 2.$

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Example 2 (Power Sum)

Evaluate $\displaystyle \lim_{n\to\infty} \frac{1^p + 2^p + \dots + n^p}{n^{p+1}}$ for $p > 0$.

Solution:
$S = \lim_{n\to\infty} \frac{1}{n} \sum_{r=1}^{n} \left( \frac{r}{n} \right)^p.$ So $f(x) = x^p$, giving: $S = \int_0^1 x^p \, dx = \left. \frac{x^{p+1}}{p+1} \right|_0^1 = \frac{1}{p+1}.$

Special cases:

• $p=1$: $\displaystyle \frac{1+2+\dots+n}{n^2} \to \frac12$.
• $p=2$: $\displaystyle \frac{1^2+2^2+\dots+n^2}{n^3} \to \frac13$.

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Example 3 (Trigonometric Sum)

Evaluate $\displaystyle \lim_{n\to\infty} \frac{1}{n} \left[ \sin\frac{\pi}{n} + \sin\frac{2\pi}{n} + \dots + \sin\frac{n\pi}{n} \right]$.

Solution:
$S = \lim_{n\to\infty} \frac{1}{n} \sum_{r=1}^{n} \sin\!\left( \frac{r\pi}{n} \right).$ So $f(x) = \sin(\pi x)$, and: $S = \int_0^1 \sin(\pi x) \, dx = \left[ -\frac{\cos(\pi x)}{\pi} \right]_0^1 = -\frac{1}{\pi} (\cos\pi - \cos 0) = \frac{2}{\pi}.$

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Example 4 (Extended Range)

Evaluate $\displaystyle \lim_{n\to\infty} \left( \frac{1}{n} + \frac{1}{n+1} + \dots + \frac{1}{3n} \right)$.

Solution:
$S = \lim_{n\to\infty} \sum_{r=0}^{2n} \frac{1}{n+r} = \lim_{n\to\infty} \sum_{r=0}^{2n} \frac{1}{n} \cdot \frac{1}{1 + \frac{r}{n}}.$ Here $\frac{r}{n}$ ranges from 0 to 2.
Thus: $S = \int_0^2 \frac{dx}{1+x} = \big[\ln(1+x)\big]_0^2 = \ln 3.$

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Example 5 (Product Inside a Limit)

Evaluate $\displaystyle \lim_{n\to\infty} \left[ \frac{(n+1)(n+2)\cdots(n+n)}{n^n} \right]^{1/n}$.

Solution:
Let $L$ be the limit. Taking logs: $\ln L = \lim_{n\to\infty} \frac{1}{n} \sum_{r=1}^{n} \ln\!\left( 1 + \frac{r}{n} \right) = \int_0^1 \ln(1+x) \, dx.$ Integrate by parts: $\int_0^1 \ln(1+x) dx = \big[ (x+1)\ln(1+x) - x \big]_0^1 = 2\ln 2 - 1.$ So $\ln L = \ln 4 - 1 = \ln\frac{4}{e}$, hence $L = \frac{4}{e}$.

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Example 6 (Rational Function)

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

Solution:
$S = \lim_{n\to\infty} \sum_{r=1}^{n} \frac{1}{n} \cdot \frac{1}{1 + \left( \frac{r}{n} \right)^2}.$ Thus $f(x) = \frac{1}{1+x^2}$, and: $S = \int_0^1 \frac{dx}{1+x^2} = \big[ \tan^{-1} x \big]_0^1 = \frac{\pi}{4}.$

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4. Previous Year JEE Problems

Problem 1 (JEE Main 2021)

Find $\displaystyle \lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^{n} \frac{2j-1+8n}{2j-1+4n}$.

Solution:
$S = \lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^{n} \frac{ \frac{2j-1}{n} + 8 }{ \frac{2j-1}{n} + 4 }.$ Let $t = \frac{2j-1}{n}$. As $j$ goes from 1 to $n$, $t$ ranges from $\frac{1}{n} \to 0$ to $\frac{2n-1}{n} \to 2$.
Thus: $S = \frac{1}{2} \int_0^2 \frac{t+8}{t+4} \, dt = \frac{1}{2} \int_0^2 \left( 1 + \frac{4}{t+4} \right) dt = \frac{1}{2} \big[ t + 4 \ln(t+4) \big]_0^2 = 1 + 2\ln\frac{3}{2}.$

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Problem 2 (JEE Advanced Pattern)

Evaluate $\displaystyle \lim_{n\to\infty} \frac{1}{\sqrt{n}} \left( \frac{1}{\sqrt{1}+\sqrt{n}} + \frac{1}{\sqrt{2}+\sqrt{n}} + \dots + \frac{1}{\sqrt{n}+\sqrt{n}} \right)$.

Solution:
$S = \lim_{n\to\infty} \sum_{r=1}^{n} \frac{1}{\sqrt{n}(\sqrt{r}+\sqrt{n})} = \lim_{n\to\infty} \frac{1}{n} \sum_{r=1}^{n} \frac{1}{ \sqrt{\frac{r}{n}} + 1 }.$ So $f(x) = \frac{1}{\sqrt{x}+1}$. Then: $S = \int_0^1 \frac{dx}{\sqrt{x}+1}.$ Substitute $\sqrt{x} = t$, $dx = 2t\,dt$: $S = \int_0^1 \frac{2t}{t+1} dt = 2 \int_0^1 \left( 1 - \frac{1}{t+1} \right) dt = 2 \big[ t - \ln(t+1) \big]_0^1 = 2(1 - \ln 2).$

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Problem 3 (JEE Advanced 2013 – Modified)

For $a \neq -1$, if
$\lim_{n\to\infty} \frac{1^a + 2^a + \dots + n^a} {(n+1)^{a-1} \big[ (na+1)+(na+2)+\dots+(na+n) \big]} = \frac{1}{60},$ find $a$.

Solution:
Numerator $\approx \frac{n^{a+1}}{a+1}$.
Denominator sum: $na+1 + na+2 + \dots + na+n = n^2a + \frac{n(n+1)}{2} \approx n^2\left(a+\frac12\right)$.
So denominator $\approx n^{a-1} \cdot n^2\left(a+\frac12\right) = n^{a+1}\left(a+\frac12\right)$.

Thus: $\frac{ \frac{n^{a+1}}{a+1} }{ n^{a+1}\left(a+\frac12\right) } = \frac{1}{(a+1)\left(a+\frac12\right)} = \frac{1}{60}.$ Hence: $(a+1)(2a+1) = 120 \quad\Rightarrow\quad 2a^2 + 3a - 119 = 0.$ Solve: $a = 7$ or $a = -\frac{17}{2}$.

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5. Common Patterns & Quick Results

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6. Tips for JEE Aspirants

1. Standard Form: Always try to rewrite the sum as $\frac{1}{n} \sum f\!\left(\frac{r}{n}\right)$.
2. Check Limits: Determine the range of $\frac{r}{n}$ carefully—it usually starts at 0 (if $r=1$) and ends at 1 (if $r=n$), but can vary.
3. Factor Out $1/n$: If a factor like $\frac{k}{n}$ appears, the integration limits may become $[0, k]$.
4. Logarithms for Products: If the limit involves a product, take logarithms to convert it into a sum.
5. Practice Common Patterns: Memorize standard results like $\ln 2$, $\pi/4$, $2/\pi$, etc.

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7. Quick‑Reference Summary

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This technique is indispensable for JEE. Practice until you can spot the conversion instantly and execute it accurately under time pressure.