Root test: Worked examples — Hinglish deep dive

4.3.13 · D3 · Maths › Calculus III — Sequences & Series › Root test

Yeh Root test ka child deep-dive hai. Yahan hum theorem re-derive nahi karte; hum har possible shape ko dhundh ke nikalte hain jo ek root-test problem le sakti hai aur har ek ke liye ek example solve karte hain. Agar exam mein koi series mile, toh woh neeche ki rows mein se kisi ek mein zaroor aayegi.

The scenario matrix

Poore note mein, $a_{n}$ ko series $\sum a_{n}$ ka $n$ -waan term maano — sum ke $n$ -wein slot mein baithne wala number ( $\sum \frac{1}{n ^{2}}$ ke liye, term $a_{n}$ hai $\frac{1}{n ^{2}}$ ). Har root-test problem ek single number se decide hoti hai: $L = lim_{n \to \infty} n ∣ a_{n} ∣ .$ Yahan $n x$ (" $x$ ka $n$ -waan root") woh number hai jo khud ko $n$ baar multiply karne par $x$ deta hai. Yeh power $n$ ko undo karta hai. Yahi reason hai ki hum ise use karte hain: agar ek term secretly $r^{n}$ ke barabar hai, toh $n r^{n} = r$ chhupa hua ratio $r$ wapas bahar le aata hai. Hum $L$ compute karte hain, phir verdict padhte hain: $L < 1$ matlab converges, $L > 1$ matlab diverges, $L = 1$ kuch nahi bolta.

Neeche ke cells har tarah ka input list karte hain jo is machine ko mil sakta hai. Example label par click karke uski full working tak jump karo.

Cell	Situation	Kya cheez ise alag banati hai	Example
A	Clean $(stuff)^{n}$ , ratio $< 1$	root power ko cancel karta hai, $L < 1$	Ex 1
B	Exponential over polynomial, base $> 1$	$L > 1$ divergence	Ex 2
C	Polynomial over exponential	polynomial root hokar $1$ banta hai, base $< 1$ bachta hai	Ex 3
D	$L = 1$ degenerate (borderline)	test chup hai, tools switch karne padenge	Ex 4
E	$L = 0$ super-fast decay	ratio effectively $0$ , strongly converges	Ex 5
F	$L = \infty$ blow-up	terms explode ho jaate hain, diverges	Ex 6
G	Negative / alternating terms	$	a_n
H	Word problem (bouncing / physical)	reality ko series mein translate karo	Ex 8
I	Exam twist: $n$ exponent ke exponent mein	$(1 + \frac{c}{n})^{n} \to e^{c}$	Ex 9

Hum sab ko cover karenge.

Cell A — clean power, ratio below 1

Worked example Ex 1 (Cell A) ·

$n = 1 \sum \infty (\frac{4 n - 1}{5 n + 2})^{n}$ Forecast: poori bracket $n$ ki power par raise ki gayi hai. Guess karo: andar ka fraction $1$ se neeche, $1$ par, ya $1$ se upar settle hoga? Padhne se pehle apna guess likh lo.

Root apply karo. $n ∣ a_{n} ∣ = n (\frac{4 n - 1}{5 n + 2})^{n} = \frac{4 n - 1}{5 n + 2} .$ Yeh step kyun? Term ek perfect $n$ -th power hai, isliye $n (\cdot)^{n} = (\cdot)$ — root aur power ek doosre ko annihilate kar dete hain. Yeh bilkul woh situation hai jiske liye root test bana hai.
Limit lo. Top aur bottom dono ko $n$ se divide karo: $\frac{4 - 1/ n}{5 + 2/ n} \to \frac{4}{5} .$ Yeh step kyun? Jab $n$ badhta hai, $1/ n \to 0$ aur $2/ n \to 0$ ; sirf leading coefficients bachte hain.
Verdict padho. $L = \frac{4}{5} < 1 \Rightarrow$ absolutely converges. Yeh step kyun? $L < 1$ ka matlab hai terms eventually ek convergent geometric series $\sum r^{n}$ se bhi tezi se shrink hoti hain jahan $\frac{4}{5} < r < 1$ .

Verify karo: $n = 100$ plug karo: $\frac{399}{502} = 0.7948 \approx 0.8$ . Trend confirm karta hai $L = 0.8 < 1$ . ✓

Cell B — exponential beats polynomial (divergence)

Cell C — polynomial over exponential (base below 1 survives)

Cell D — the silent case $L = 1$

Worked example Ex 4 (Cell D) ·

$n = 1 \sum \infty \frac{1}{n ( ln n ) ^{2}}$ ( $n = 2$ se start) Forecast: yahan koi obvious $n$ -th power nahi hai. Guess karo ki root test yeh decide bhi kar payega ya nahi.

Root compute karo. $n \frac{1}{n ( ln n ) ^{2}} = \frac{1}{n ^{1/ n} ( ln n ) ^{2/ n}} .$ Yeh step kyun? Har factor ka root lo: $n n = n^{1/ n}$ aur $n (ln n)^{2} = (ln n)^{2/ n}$ .
Dono roots $\to 1$ . $n^{1/ n} \to 1$ ; aur $(ln n)^{2/ n} = e^{\frac{2 l n ( l n n )}{n}} \to e^{0} = 1$ kyunki $\frac{l n ( l n n )}{n} \to 0$ . Yeh step kyun? Jo bhi cheez $n$ mein exponential se dheemar badhti hai, woh root hokar $1$ banti hai. Dono factors qualify karte hain.
Verdict. $L = 1 \Rightarrow$ inconclusive — root test decide nahi kar sakta. Yeh step kyun? $L = 1$ blind spot hai: yeh convergent aur divergent tails mein fark nahi kar sakta.
Tools switch karo. Comparison test integral $\int_{2}^{\infty} \frac{d x}{x ( l n x ) ^{2}} = [- \frac{1}{l n x}]_{2}^{\infty} = \frac{1}{l n 2}$ (finite) ke saath dikhata hai ki series converges.

Verify karo: antiderivative check: $\frac{d}{d x} (- \frac{1}{l n x}) = \frac{1}{x ( l n x ) ^{2}}$ . ✓ Aur $L = 1$ exactly boundary par hai. ✓

Cell E — decay so fast that $L = 0$

Cell F — blow-up, $L = \infty$

Cell G — negative / alternating terms

Cell H — the word problem

Worked example Ex 8 (Cell H) · Ek bouncing signal

Ek radar pulse har hop mein power khoti hai. Hop $n$ par woh $E_{n} = (\frac{3}{4})^{n} n^{2}$ joules energy carry karta hai. Kya sabhi hops par total energy ek finite amount mein converge karti hai? Forecast: $\frac{3}{4}$ ise shrink karta hai, lekin $n^{2}$ push back karta hai. Kaun jeeta hai?

Series ke roop mein model karo. Total energy $= \sum_{n = 1}^{\infty} E_{n} = \sum n^{2} (\frac{3}{4})^{n} .$ Yeh step kyun? "Sabhi hops par total" literally har hop ki energy ka infinite sum hai.
Har factor ka root lo. $n n^{2} (\frac{3}{4})^{n} = (n n)^{2} \cdot \frac{3}{4} \to 1 \cdot \frac{3}{4} = \frac{3}{4} .$ Yeh step kyun? Polynomial $n^{2}$ root hokar $1$ banta hai (Ex 2 wali logic); geometric base $\frac{3}{4}$ bachta hai.
Verdict. $L = \frac{3}{4} < 1 \Rightarrow$ converges: total energy finite hai. Physical geometric decay $\frac{3}{4} < 1$ polynomial growth ko haara deta hai.

Verify karo (units + numerics): $E_{n}$ joules mein; joules ka sum joules hota hai — dimensionally theek hai. Terms: $E_{1} = 0.75, E_{10} \approx 5.6, E_{30} \approx 0.16, E_{50} \approx 0.001$ — pehle peak phir $0$ ki taraf decay, total finite. ✓

Neeche ka figure yeh competition visible banata hai: cyan bars per-hop energies $E_{n}$ hain (woh ek early peak tak uthte hain, phir khatam ho jaate hain), jabki amber curve running total hai — dekho yeh ek finite plateau par flat hota hai, jo exactly wahi hai jo $L = \frac{3}{4} < 1$ guarantee karta hai.

Cell I — exam twist exponent inside an exponent

Recall Matrix par khud ko test karo

Kaun sa cell: $\sum (n / (3 n + 1))^{n}$ ? ::: Cell A — clean power, $L = 1/3 < 1$ , converges. Kaun sa cell: $\sum 7^{n} / n^{10}$ ? ::: Cell B — exponential over polynomial, $L = 7 > 1$ , diverges. Kaun sa cell: $\sum 1/ n^{p}$ (koi bhi $p$ )? ::: Cell D — $L = 1$ , inconclusive; p-series use karo. Kaun sa cell: $\sum 1/ (2 n)^{n}$ ? ::: Cell E — $L = 0$ , strongly converges. Kya alternating sign $L$ change karta hai? ::: Nahi — root test $∣ a_{n} ∣$ use karta hai, isliye sign mit jaati hai (Cell G).