4.3.7 · D1Integral test — proof, p-series
Chhote positive numbers ki kabhi na khatam hone wali sum asal mein sirf ek staircase of rectangles hai, aur woh smooth curve jise ye rectangles hug karte hain woh kuch aisa hai jo hum pehle se measure karna jaante hain: area . Agar hum area measure kar sakein, toh hum decide kar sakte hain ki staircase sky tak pahunchti hai ya kisi finite height par ruk jaati hai.
the Integral Test note padhne se pehle, tumhe us mein aane wala har symbol properly samajhna hoga. Yeh page unhe order mein build karta hai, har ek us picture se jo woh represent karta hai. Yahan kuch bhi assume nahi kiya gaya ki tumne pehle calculus notation dekhi hai line one se pehle.
::: self-test lines kaise padhein
Is poore page mein tumhe chhoti lines milein gi jo Question ::: Answerflip-card prompts hain: teen colons se pehle wala part padho, apne mann mein jawab do, phir teen colons ke baad wale part se check karo. Teen colons sirf question aur hidden answer ke beech ka divider hain.
Definition Sequence aur terma n sequence numbers ki ek ordered list hoti hai, har ek counting number 1 , 2 , 3 , … n a n a-sub-n ". Neeche ka chhota n slot number hai.
Picture: socho ek row mein numbered mailboxes hain. Box 1 a 1 2 a 2
Intuition Topic ko yeh kyun chahiyePoora subject in box-contents ko ek saath add karne ke baare mein hai. Toh pehle humein ek clean naam chahiye — a n n
a n List 1 , 2 1 , 3 1 , … a n = n 1 a 5 = 5 1
∑ ∑ "inhe add karo" ka shorthand hai. Neeche hum likhte hain ki slot number kahan se shuru hota hai, upar likhte hain kahan khatam hota hai.
∑ n = 1 4 a n = a 1 + a 2 + a 3 + a 4 .
YEH kya kehta hai: n = 1 a n n = 4
Definition Ek infinite seriesJab sigma ke upar sideways-eight ∞ infinity — "chalti rehti hai, koi last number nahi") hoti hai, toh matlab hai hum kabhi add karna band nahi karte:
∑ n = 1 ∞ a n = a 1 + a 2 + a 3 + ⋯ series kehte hain.
∞ fate decide ki jaaye bina woh impossible addition kiye. Toh symbol ∞
∑ n = 1 3 n 1 1 + 2 1 + 3 1 = 6 11
S N slot N aur woh finite total dekhte hain:
S N = ∑ n = 1 N a n = a 1 + a 2 + ⋯ + a N . N N
Picture: dekho total kaise badhta hai jab tum har baar ek aur box add karte ho. S 1 , S 2 , S 3 , …
Definition Sequence ka convergence,lim S 1 , S 2 , S 3 , … L lim N → ∞ S N = L , S N aur bani rehti hain single finite number L N S N L L nahi milta (values bina bound ke badhti hain ya kabhi settle nahi karti), toh sequence diverge karti hai.
Series ka convergence vs divergence
Series ko poori tarah se uske partial sums ke through judge kiya jaata hai:
Series ∑ a n converges exactly tab jab N → ∞ lim S N = L L L define karte hain.
Series diverges tab jab partial-sum sequence S N
Intuition Partial sums kyun real definition hain"Kya infinite sum settle hoti hai?" literally yeh sawaal hai "kya lim N → ∞ S N L S N S N
n ∑ a n zaroor zero tak shrink hone chahiye :
lim n → ∞ a n = 0. lim n → ∞ a n = 0 turant diverge karti hai — aage koi kaam nahi. Yeh n-th term test for divergence hai.
Intuition Necessary, sufficient nahiTerms ka zero par jaana ek darwaza hai jisse tumhe guzarna hoga, arrival ki guarantee nahi. Isse guzarna (a n → 0 permission le rahe ho jiska jawab integral test deta hai. Fail karna story wahan khatam karta hai. Harmonic series mein a n = n 1 → 0
a n → 0 Yeh necessary hai sufficient nahi; ∑ n 1 a n → 0
Definition Increasing aur bounded above
Ek sequence increasing hoti hai agar har term pehle wali se ≥ S 1 ≤ S 2 ≤ S 3 ≤ ⋯
Woh bounded above hai agar ek ceiling number M S N ≤ M N
Sequence ka least upper bound (supremum) sabse chhota aisa ceiling hai — sabse low number L S N ≤ L N
Agar a n positive hain, toh har step mein positive amount add karne ka matlab hai partial sums sirf upar jaenge — woh automatically increasing hain.
Intuition Monotone ceiling pictureSocho running totals ek ladder chadhte hue, kabhi neeche nahi aate. Upar se ek ceiling M lowest height ke neeche pile up karna hi padega jo koi kabhi cross nahi karta (least upper bound). Woh "zaroor settle hoga" fact exactly wahi hai jo integral test apne final step mein use karta hai: integral ceiling M
"Positive terms" kyun matter karta hai Positive terms ke saath totals sirf badhte hain, toh ek ceiling convergence force karti hai; mixed signs ke saath totals wobble kar sakte hain aur yeh reasoning toot jaati hai.
Definition Function aur uski curveEk function f x f ( x ) graph woh smooth curve hai jo tumhe milti hai jab tum har input ke liye point ( x , f ( x )) x hi output f ( x )
[ 1 , ∞ ) f domain [ 1 , ∞ ) x 1 [ 1 included hai; "∞ ) N [ N , ∞ ) [ 1 , ∞ )
Sequences se bridge: agar hum a n = f ( n ) x = 1 , 2 , 3 , …
Intuition Sum ko function mein kyun badleinEk jagged sum mushkil hai; ek smooth curve ka ek tool hai jo hum jaante hain — area . a n f ( n )
Definition Positive, continuous, decreasing (on[ 1 , ∞ ) f [ 1 , ∞ )
Positive : curve x upar rehti hai, f ( x ) > 0 Continuous : curve mein koi break ya jump nahi — toh uske neeche area actually exist karta hai.Decreasing : curve right move karne par girti hai, toh baad ka input kabhi taller output nahi deta. Iska matlab hai har width-1 left edge sabse tall point hai aur right edge sabse chhota. Kyunki har strip ek taraf jhukti hai, rectangles ko area se ek consistent direction mein compare kiya ja sakta hai — left-edge rectangles hamesha curve ke upar baithte hain, right-edge rectangles hamesha neeche.
∫ ∫ a b f ( x ) d x exact area hai jo curve y = f ( x ) x x = a x = b ∫ d x x
Intuition Area hamare sawaal ka jawab kyun deta haiSeries ke rectangles aur curve ke neeche area itne close hain ki woh ek doosre ko trap karte hain. Agar smooth area finite hai, toh boxed rectangles us finite ceiling se zyada total nahi kar sakte — series converge karti hai. Agar area infinite hai, toh rectangles uske saath upar khichwate hain. Area sum ki fate decide karta hai. Yeh poore method ka dil hai.
Definition Improper integral aur limitlim Infinity tak area measure karne ke liye hum ∞ b b ∫ 1 ∞ f ( x ) d x = lim b → ∞ ∫ 1 b f ( x ) d x . lim b → ∞ b converges ; agar hamesha ke liye badhe toh diverges . Dekho Improper integrals .
Yahan limit kyun chahiye Hum kabhi "infinity tak pahunch" nahi sakte, toh hum area ko wall b
p n p 1 p power hai jis par slot number raise kiya jaata hai. Ek negative exponent sirf "one over" matlab hai: x − p = x p 1 p faster shrink karta hai, jo exactly woh knob hai jo convergence decide karta hai.
Definition Natural logarithmln ln x special number e ≈ 2.718 x " Yeh isliye aata hai kyunki curve y = 1/ x ln x
x − 2 x 2 1 x
ln y = 1/ x 1 b ln b ln b → ∞ ∑ n 1
Neeche diagram (jahan Mermaid available ho wahan render hota hai) build order dikhata hai; agar plain text mein dikhe, toh ise top-to-bottom padho jaise §1–§8 mein describe ki gayi same dependency chain hai.
Monotone convergence theorem
Function f x and its graph
Positive continuous decreasing
Improper integral with limit
Khud test karo — sirf jawab dene ke baad reveal karo.
a n Woh number jo ek ordered list ke slot n
∑ n = 1 ∞ a n Har list-term ko hamesha ke liye add karo, slot 1 se shuru karke.
Partial sum S N Finite running total a 1 + ⋯ + a N N
Series ke convergence ko limit notation mein likho. ∑ a n lim N → ∞ S N = L L L
Woh n Agar ∑ a n lim n → ∞ a n = 0 a n → 0
Monotone Convergence Theorem batao. Ek increasing sequence jo bounded above ho woh converge karti hai, apne least upper bound (supremum) par.
"Increasing aur bounded above" convergence kyun force karta hai? Badhte totals jo ceiling se block hon unhe apne least upper bound ke neeche pile up karna hi padta hai.
f [ 1 , ∞ )
f Positive (koi cancelling nahi), continuous (area exist karta hai), decreasing (clean rectangle comparison).
∫ 1 b f ( x ) d x Curve aur x b
lim b → ∞ Right wall ko infinity tak slide karta hai aur woh number report karta hai jiske pass area jaata hai.
x − p 1/ x p
ln x e ≈ 2.718 x
Integral test — proof, p-series (index 4.3.7) — woh topic jise yeh foundations unlock karte hain.Improper integrals — §7 ki limit-to-infinity area machine.Harmonic series — jahan §8 ki slow ln n-th term test for divergence — §4 ka necessary (sufficient nahi) precondition.Comparison test · Limit comparison test — sibling tests jo usi a n = f ( n ) Riemann zeta function — ∑ 1/ n p 
