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PUNE MCA Exam 2014 Question Paper with answers for online practice

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Question : 26 of 58

Marks: +1, -0

Evaluate limit

\lim\limits_{N\to x}

n sin(2πn!e) equal

π
$\frac{\pi}{2}$
2π
None of these

Solution:

Concept:

Evaluate

\lim_{n\to\infty} n \sin(2\pi n! e)

by expanding

e

as a series.

Explanation:

First, note

e = \sum_{k=0}^{\infty} \frac{1}{k!}

.

For large

n

,

n!e = n! \sum_{k=0}^{\infty} \frac{1}{k!} = I_n + f_n

, where

I_n

is an integer (since terms up to

1/n!

yield integers when multiplied by

n!

) and

f_n = \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \cdots

is the fractional part.

Thus

\sin(2\pi n! e) = \sin(2\pi I_n + 2\pi f_n) = \sin(2\pi f_n)

.

For large

n

,

f_n \sim \frac{1}{n+1}

, so

\sin(2\pi f_n) \sim 2\pi f_n

.

Then

n \sin(2\pi n! e) \sim n \cdot 2\pi f_n = 2\pi n f_n

.

Now

n f_n = n\left( \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \cdots \right) = \frac{n}{n+1} + \frac{n}{(n+1)(n+2)} + \cdots

.

As

n\to\infty

,

\frac{n}{n+1}\to 1

and all subsequent terms tend to

0

. Hence

n f_n \to 1

, so the limit is

2\pi

.

Answer:

2\pi

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