#3. Thanks. 3. If an1000"n!Find N E I, the set of positive integers, such that an+1 < an(n > N).

Given an=1000nn!. We have to find the N∈I such that an+1<an, n>N. For n=1 a1=100011! For n=2…

Answered: 3. If а, 1000". Find N E I, the set of…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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#3. Thanks.

3. If an
1000"
n!
Find N E I, the set of positive integers, such that an+1 < an
(n > N).

Expert Solution

Step 1

Given $a_{n} = \frac{1000^{n}}{n!}$ .

We have to find the $N \in I$ such that $a_{n + 1} < a_{n}$ , $(n > N)$ .

For $n = 1$

$a_{1} = \frac{1000^{1}}{1!}$

For $n = 2$

$a_{2} = \frac{1000^{2}}{2!} = \frac{1000}{2} \cdot \frac{1000^{1}}{1!} = \frac{1000}{2} \cdot a_{1}$

For $n = 3$

$a_{3} = \frac{1000^{3}}{3!} = \frac{1000}{3} \cdot \frac{1000^{2}}{2!} = \frac{1000}{3} \cdot a_{2}$

Similarly, $a_{4} = \frac{1000}{4} \cdot a_{3}$

$a_{5} = \frac{1000}{5} \cdot a_{4}$

$⋮$

$a_{1000} = \frac{1000}{1000} \cdot a_{999} = a_{999}$

$a_{1001} = \frac{1000}{1001} \cdot a_{1000}$

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3. If а, 1000". Find N E I, the set of positive integers, such that an+1 < an (n > N). n!