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Show that nnnlimn→∞11.2+12.3+13.4+...+1n(n+1) = 1 - Mathematics

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प्रश्न

Show that `lim_("n" -> oo) 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/("n"("n" + 1))` = 1

योग
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उत्तर

Tn = `1/("n"("n" + 1))`

`1/("n"("n" + 1)) = "A"/"n" + "B"/("n" +1)`

1 = A(n + 1) + Bn

Put n = 0

1 = A(0 + 1) + B × 0

A = 1

Put n = – 1

1 = A(– 1 + 1) + B(– 1)

1 = 0 – B

⇒ B = – 1

`1/("n"("n" + 1)) = 1/"n" - 1/("n" + 1)`

Tn = `1/"n" - 1/("n" + 1)`

T1 = `1/1- 1/2`

T2 = `1/2 - 1/3`

T3 = `1/3 - 1/4`

........................
........................

Tn = `1/"n" - 1/("n" + 1)`

T1 + T2 + .... + Tn = `(1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + .... + (1/"n" - 1/("n" + 1))`

T1 + T2 + .... + Tn = `(1 - 1/("n" + 1))`

`lim_("n" -> oo) ("T"_1 + "T"_2 + .... + "T"_"n") = lim_("n" -> oo) (1 - 1/("n" + 1))`

= `1 - lim_("n" -> oo) 1/("n" + 1)`

= 1 – 0

 `lim_("n" -> oo) 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/("n"("n" + 1))` = 1

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अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.3 [पृष्ठ १११]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.3 | Q 8. (iii) | पृष्ठ १११

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