Integral test — proof, p-series

4.3.7 · Maths › Calculus III — Sequences & Series

Tag: #calculus3 #series/convergence

The big question (WHY)

Hamare paas ek series $\sum a_n$ hai aur hum jaanna chahte hain: kya yeh infinite sum kisi finite number par settle hoti hai, ya blow up ho jaati hai? Infinitely many terms ko haath se add karna impossible hai. Integral test ek trick hai: discrete sum (rectangles ki ek staircase) ko ek continuous integral (curve ke neeche ka area) se replace karo jo hum actually compute kar sakte hain.

The test itself (WHAT)

Teeno conditions matter karti hain: positive (taaki areas cancel na hon), decreasing (taaki rectangle comparison ek saaf direction mein jaaye), continuous (taaki integral exist kare).

Deriving it from scratch (HOW)

Hum woh inequality banate hain jo sum ko trap karti hai. Socho $y=f(x)$ daayein taraf girta ja raha hai.

Step 1 — Upper-sum rectangles (left endpoints). Har interval $[n,n+1]$ par, kyunki $f$ decrease karta hai, sabse badi value left par hoti hai, $f(n)$, aur sabse choti right par, $f(n+1)$. Width $1$ aur height $f(n)$ (left endpoint) wala ek rectangle banao: yeh curve ke upar baithta hai, isliye iska area upper sum hai. Yeh step kyun? Left-endpoint height $f(n)$ interval $[n,n+1]$ par area ko overestimate karta hai, jo hume integral par ek upper bound deta hai.

${\int }_n^{n+1}f(x)dx\le f(n).$

$n=1$ se $N$ tak sum karo: ${\int }_1^{N+1}f(x)dx\le {\sum }_{n=1}^{N}f(n)=S_N.$

Step 2 — Lower-sum rectangles (right endpoints). Ab height $f(n+1)$ (right endpoint) use karo, jo curve ke neeche baithta hai, isliye iska area lower sum hai: $f(n+1)\le {\int }_n^{n+1}f(x)dx.$ $n=1$ se $N-1$ tak sum karo: $S_N-f(1)={\sum }_{n=2}^{N}f(n)\le {\int }_1^{N}f(x)dx.$

Step 3 — Squeeze. Combine karo: ${\int }_1^{N+1}f(x)dx\le S_N\le f(1)+{\int }_1^{N}f(x)dx.$

Step 4 — Conclusion nikalo.

• Agar ${\int }_1^{\infty }f$ converges (finite), toh right side bounded rehta hai, isliye $S_N$ increasing aur upar se bounded hai ⇒ $\sum a_n$ converge karta hai.
• Agar ${\int }_1^{\infty }f$ diverges ($\to \infty$), toh left side $\to \infty$, aur $S_N\to \infty$ bhi ho jaata hai ⇒ $\sum a_n$ diverge karta hai. $■$

The payoff: the p-series (80/20)

Test apply karo $f(x)=x^{-p}$ ke saath (positive, continuous, $p>0$ ke liye decreasing). Integral compute karo.

Case $p\ne 1$: ${\int }_1^{\infty }{x}^{-p}dx={\lim }_{b\to \infty }\frac{x^{1-p}}{1-p}{|}_1^b={\lim }_{b\to \infty }\frac{b^{1-p}-1}{1-p}.$

• Agar $p>1$: $1-p<0$, isliye $b^{1-p}\to 0$, integral $=\frac{1}{p-1}$ (finite) ⇒ converges.
• Agar $p<1$: $1-p>0$, isliye $b^{1-p}\to \infty$ ⇒ diverges.

Case $p=1$ (harmonic series): ${\int }_1^{\infty }\frac{dx}{x}={\lim }_{b\to \infty }\ln b=\infty \Rightarrow \text{diverges}.$

Worked examples

Steel-manning the classic mistakes

Active recall

Connections

• Comparison test — p-series ko standard "yardstick" ki tarah use karta hai.
• Limit comparison test — $1/n^p$ ke against compare karo.
• Improper integrals — proof ke andar ka engine.
• Harmonic series — boundary case $p=1$.
• Riemann zeta function — $\zeta (p)=\sum 1/n^p$ for $p>1$.
• n-th term test for divergence — necessary (sufficient nahi) condition.

Concept Map