NIMCET 2012 Question Paper with solutions

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Question : 6 of 120

Marks: +1, -0

The value of

\lim↙{n→∞} π/2 [sinπ/2+sin{2π}/n+...+sin{(n-1)π}/n]

is

0
π
2
$π/2$

Solution:

= \lim↙{n→∞} π/n \{sin(π/n) + sin(π/n+π/n) + sin (π/n+{2π}/n) + ... + sin (π/n+{nπ}/n)\}

Since ,

sin α + sin (α + β) + sin (α + 2β) + ... + sin (α + nβ)

=

{sin({2α+nβ}/2)sin{nβ}/2}/{sinβ/2}

=

\lim↙{n→∞} π/n {sin\{π/n+(π/n+{nπ}/n)\}sinn/2π/n}/{sinπ/{2n}}

=

\lim↙{n→∞} π/n {sin({2π+nπ}/n)sinπ/2}/{sinπ/{2n}}

=

\lim↙{n→∞} 1/{2({sinπ/{2n}}/{π/{2n}})} . \sin (π-{2π}/n) . 1

Since

\lim↙{θ→∞} {sin1/θ}/{1/θ} = 1

=

1/{2} . 1 . sin (π + 0)

=

1/2 . 0 = 0

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