Root test: Exercises — Hinglish deep dive

4.3.13 · D4 · Maths › Calculus III — Sequences & Series › Root test

Level 1 — Recognition

Goal: decide karo ki root test sahi tool bhi hai ya nahi, aur $L$ read off karo jab term ek clean $n$ -th power ho.

Recall Solution 1.1

Poori term $n$ -th power mein raised hai $(\dots)^{n}$ . Yahi root test ka trigger hai: $n \cdot$ outer power ko exactly cancel kar dega. $n (\frac{n}{2 n + 1})^{n} = \frac{n}{2 n + 1} .$ Humne kya kiya / kyun: $n$ -th root lena " $n$ -th power mein raise karna" undo karta hai, base bacha deta hai. Ab top aur bottom ko $n$ se divide karo: $\frac{n}{2 n + 1} = \frac{1}{2 + 1/ n} n \to \infty \frac{1}{2} .$ To $L = \frac{1}{2} < 1$ $\Rightarrow$ converges absolutely.

Recall Solution 1.2

Har factor ka alag root lo: $n \frac{5 ^{n}}{n ^{4}} = \frac{n 5 ^{n}}{n n ^{4}} = \frac{5}{( n n ) ^{4}} .$ Yeh step kyun: $n 5^{n} = 5$ (ek $n$ -th power ka $n$ -th root), aur $n n^{4} = (n^{1/ n})^{4}$ . Kyunki $n n \to 1$ , denominator $\to 1^{4} = 1$ , isliye $L = \frac{5}{1} = 5 > 1 \Rightarrow diverges .$ Upar wala exponential $5^{n}$ polynomial $n^{4}$ ko beat kar deta hai — polynomial factor root test ko visible nahi hota.

Recall Solution 1.3

Denominator mein perfect $n$ -th power hai: $n \frac{1}{( a r c t a n n ) ^{n}} = \frac{1}{a r c t a n n} .$ Yeh step kyun: $n (arctan n)^{n} = arctan n$ kyunki base exponent ke relative fixed hai. Jaise $n \to \infty$ , $arctan n \to \frac{π}{2} \approx 1.5708$ , isliye $L = \frac{1}{π /2} = \frac{2}{π} \approx 0.6366 < 1 \Rightarrow converges .$

Level 2 — Application

Goal: standard limit $n n \to 1$ aur algebra use karo $L$ get karne ke liye jab term powers ka mix ho.

Recall Solution 2.1

$n \frac{n ^{10}}{3 ^{n}} = \frac{( n n ) ^{10}}{3} \to \frac{1 ^{10}}{3} = \frac{1}{3} < 1.$ Kya/kyun: root ko top aur bottom par split karo; top hai $(n^{1/ n})^{10} \to 1$ , bottom ek clean $3$ hai. $L = \frac{1}{3}$ $\Rightarrow$ converges.

Recall Solution 2.2

Exponent $n^{2}$ hai, $n$ nahi — lekin root test phir bhi shine karta hai kyunki $n (\dots)^{n^{2}} = (\dots)^{n}$ : $n (1 - \frac{1}{n})^{n^{2}} = (1 - \frac{1}{n})^{n} .$ Yeh step kyun: $n x^{n^{2}} = x^{n^{2} / n} = x^{n}$ . Ab famous limit use karo $(1 - \frac{1}{n})^{n} \to e^{- 1} .$ Yeh limit kyun: $(1 + \frac{c}{n})^{n} \to e^{c}$ ; yahan $c = - 1$ . To $L = e^{- 1} = \frac{1}{e} \approx 0.3679 < 1 \Rightarrow converges .$

Recall Solution 2.3

$n \frac{2 ^{n} n ^{3}}{5 ^{n}} = \frac{2 ( n n ) ^{3}}{5} \to \frac{2 \cdot 1}{5} = \frac{2}{5} < 1.$ Kya/kyun: do exponentials apni bases $2$ aur $5$ par root karte hain; polynomial $(n^{1/ n})^{3} \to 1$ . $L = \frac{2}{5}$ $\Rightarrow$ converges.

Recall Solution 2.4

$n (\frac{3 n ^{2} + 1}{n ^{2} + 4})^{n} = \frac{3 n ^{2} + 1}{n ^{2} + 4} \to 3.$ Yeh step kyun: top aur bottom ko $n^{2}$ se divide karo: $\frac{3 + 1/ n ^{2}}{1 + 4/ n ^{2}} \to 3$ . Kyunki $L = 3 > 1$ $\Rightarrow$ diverges.

Level 3 — Analysis

Goal: $L = 1$ boundary spot karo, aur jaano ki root test kab jawab dene se mana karta hai.

Recall Solution 3.1

$n \frac{1}{n ( l n n ) ^{2}} = \frac{1}{n ^{1/ n} ( l n n ) ^{2/ n}} .$ Har piece kyun: $n^{1/ n} \to 1$ (standard limit), aur $(ln n)^{2/ n} = e^{\frac{2 l n l n n}{n}} \to e^{0} = 1$ kyunki $\frac{l n l n n}{n} \to 0$ ( $ln ln n$ bhi $n$ se kaafi dheeray badhta hai). Isliye $L = \frac{1}{1 \cdot 1} = 1 \Rightarrow inconclusive .$ Aage kya: tools switch karo — integral test dikhata hai ki yeh series converges karti hai, lekin root test ise simply dekh nahi sakta.

Recall Solution 3.2

$n \frac{n !}{n ^{n}} = \frac{( n ! ) ^{1/ n}}{n} .$ Yeh subtle kyun hai: factorial $n!$ ek clean power nahi hai. Known asymptotic use karo $(n!)^{1/ n} \sim \frac{n}{e}$ (Stirling se: $n! \sim 2 π n (n / e)^{n}$ , to $(n!)^{1/ n} \sim n / e$ ). Tab $L = \frac{( n ! ) ^{1/ n}}{n} \to \frac{n / e}{n} = \frac{1}{e} \approx 0.3679 < 1 \Rightarrow converges .$ Comment: Ratio test yahan aur bhi clean hai — $\frac{a _{n + 1}}{a _{n}} = (\frac{n}{n + 1})^{n} \to e^{- 1}$ , same $L$ . Isliye ye dono tests cousins hain: bahut si series ke liye dono tumhe same number dete hain.

Recall Solution 3.3

$n \frac{1}{n ^{1 + 1/ n}} = n^{- (1 + 1/ n) / n} = e^{- \frac{( 1 + 1/ n ) l n n}{n}} .$ Exponential form kyun: moving exponent wali power ki limit lene ke liye, $x^{y} = e^{y l n x}$ likhte hain. Exponent hai $- \frac{( 1 + 1/ n ) l n n}{n} \to - 1 \cdot 0 = 0,$ kyunki $\frac{l n n}{n} \to 0$ . To $L = e^{0} = 1 \Rightarrow$ inconclusive. (Comparison dikhata hai ki yeh actually diverges karta hai, $\sum 1/ n$ ke close hai.)

Level 4 — Synthesis

Goal: root test ko power-series ideas ke saath combine karo aur root aur ratio mein se choose karo.

Recall Solution 4.1

Yahan $a_{n} = \frac{x ^{n}}{n 2 ^{n}}$ . Root lo: $n ∣ a_{n} ∣ = \frac{∣ x ∣}{2 ( n n )} \to \frac{∣ x ∣}{2} .$ Kyun: $∣ x ∣^{n}$ root hokar $∣ x ∣$ ban jaata hai, $2^{n}$ root hokar $2$ ban jaata hai, aur $n n \to 1$ . Convergence ke liye $L < 1$ chahiye: $\frac{∣ x ∣}{2} < 1 ⟺ ∣ x ∣ < 2.$ To radius of convergence hai $R = 2$ (compare Radius of convergence, Cauchy–Hadamard). Endpoints alag test karo (root test wahan silent hai kyunki $L = 1$ ):

$x = 2$ : $\sum \frac{2 ^{n}}{n 2 ^{n}} = \sum \frac{1}{n}$ diverges.
$x = - 2$ : $\sum \frac{( - 2 ) ^{n}}{n 2 ^{n}} = \sum \frac{( - 1 ) ^{n}}{n}$ converges conditionally (alternating harmonic — alternating series test convergence deta hai, lekin $\sum \frac{1}{n}$ dikhata hai ki yeh absolutely convergent nahi hai, aur precisely isliye root test, jo ek absolute-convergence test hai, yahan silent rehta hai).

Interval of convergence: $[- 2, 2)$ .

Recall Solution 4.2

Exponent $n^{2}$ root test ko natural banata hai (yeh $n^{2}$ ko $n$ mein turn karta hai): $n (\frac{n}{n + 1})^{n^{2}} = (\frac{n}{n + 1})^{n} = (1 - \frac{1}{n + 1})^{n} .$ Kyun: $n x^{n^{2}} = x^{n}$ . Jaise $n \to \infty$ , $(1 - \frac{1}{n + 1})^{n} \to e^{- 1}$ (same $(1 + \frac{c}{n})^{n} \to e^{c}$ idea, $c = - 1$ ). To $L = e^{- 1} = \frac{1}{e} \approx 0.3679 < 1 \Rightarrow converges .$ Ratio test ke liye tumhe $(\frac{n + 1}{n + 2})^{(n + 1)^{2}} / (\frac{n}{n + 1})^{n^{2}}$ simplify karna padta — algebra hell. Root jeet jaata hai.

Recall Solution 4.3

$n (\frac{p n}{n + 1})^{n} = \frac{p n}{n + 1} \to p .$ Kyun: $\frac{p n}{n + 1} = \frac{p}{1 + 1/ n} \to p$ . To $L = p$ .

$p < 1$ : $L < 1$ $\Rightarrow$ converges.
$p > 1$ : $L > 1$ $\Rightarrow$ diverges.
$p = 1$ : $L = 1$ , root test inconclusive. Tab $a_{n} = (\frac{n}{n + 1})^{n} \to e^{- 1} \neq = 0$ , to n-th term divergence test se yeh diverges.

Answer: converges iff $0 < p < 1$ .

Level 5 — Mastery

Goal: limsup handle karo (ordinary limit exist nahi karta), aur structural facts prove karo.

Recall Solution 5.1

$n a_{n}$ ki do subsequential values compute karo:

even $n$ : $n 2^{- n} = \frac{1}{2}$ ,
odd $n$ : $n 3^{- n} = \frac{1}{3}$ .

Limsup kyun: sequence $n a_{n}$ $\frac{1}{2}$ aur $\frac{1}{3}$ ke beech oscillate karta hai, isliye koi single limit exist nahi karta. $lim sup$ sabse bada subsequential limit hota hai: $L = lim sup_{n \to \infty} n a_{n} = max (\frac{1}{2}, \frac{1}{3}) = \frac{1}{2} < 1.$ $L = \frac{1}{2} < 1 \Rightarrow$ converges absolutely. (Yahi reason hai ki parent note $L$ ko $lim sup$ se define karta hai — taaki test ka hamesha ek value ho, jab plain limit fail bhi kare.)

Recall Solution 5.2

Bade $n$ ke liye denominator $3^{n}$ se dominate hota hai ( $n^{3}$ negligible hai). Isko root karo: $n \frac{2 ^{n}}{3 ^{n} + n ^{3}} = \frac{2}{n 3 ^{n} + n ^{3}} .$ Yeh step kyun: denominator se largest term factor out karo: $3^{n} + n^{3} = 3^{n} (1 + \frac{n ^{3}}{3 ^{n}})$ , to $n 3^{n} + n^{3} = 3 (1 + \frac{n ^{3}}{3 ^{n}})^{1/ n} .$ Bracket ka root $\to 1$ kyun: $ε_{n} = \frac{n ^{3}}{3 ^{n}}$ likho, jo $\to 0$ jaata hai (exponential polynomial ko beat karta hai). Tab $(1 + ε_{n})^{1/ n} = e^{\frac{l n ( 1 + ε _{n} )}{n}},$ aur exponent $\frac{l n ( 1 + ε _{n} )}{n} \to \frac{l n ( 1 + 0 )}{\infty} = \frac{0}{\infty} = 0$ , to bracket ka root $\to e^{0} = 1$ . Isliye $L = \frac{2}{3 \cdot 1} = \frac{2}{3} < 1 \Rightarrow converges .$

Recall Solution 5.3

Step 1 — square ka root. Har $n$ ke liye, $n ∣ a_{n} ∣^{2} = (n ∣ a_{n} ∣)^{2} .$ Kyun: $∣ a_{n} ∣^{2} = (∣ a_{n} ∣)^{2}$ , aur $n$ -th root lena squaring ke saath commute karta hai: $(x)^{2/ n} = (x^{1/ n})^{2}$ . Step 2 — limit ko square ke through pass karo. Squaring continuous hai, isliye $lim_{n} (n ∣ a_{n} ∣)^{2} = (lim_{n} n ∣ a_{n} ∣)^{2} = L^{2} .$ Step 3 — conclusion. Agar $L < 1$ to $L^{2} < 1$ bhi ( $[0, 1)$ mein kisi number ko square karna use $[0, 1)$ mein hi rakhta hai). $\sum a_{n}^{2}$ par root test apply karne se, woh series bhi converges absolutely. To root test ke through absolute convergence squared series mein inherit hoti hai. $■$

Recall Self-test checklist (grade karne ke liye reveal karo)

Root $n$ -th power cancel karta hai ::: haan — $n (\dots)^{n} = (\dots)$ Polynomial factor $n^{k}$ root hokar banta hai ::: $1$ $L = 1$ ka matlab hai ::: inconclusive, tools switch karo Power-series interval ke endpoints ko chahiye ::: alag manual testing Jab plain limit fail kare, use karo ::: $lim sup$ (sabse bada subsequential limit)