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Given [tex] a_n = \frac {n}{\alpha^n} [/tex] and [tex] \alpha [/tex] is a number greater than 1, we assert as n increases the sequence of numbers [tex] a_n = \frac {n}{\alpha^n} [/tex] tends to the limit 0.

Let us consider the sequence [tex] \sqrt a_n = \frac {\sqrt n}{(\sqrt \alpha)^n}[/tex]

We put [tex] \sqrt \alpha = 1+h [/tex] where [tex] h > 0 [/tex]

[tex] \sqrt \alpha^n = (1+h)^n > 1+nh [/tex]

[tex]\sqrt a_n = \frac {\sqrt n}{(1+h)^n} \leq \frac {\sqrt n}{1+nh} \leq \frac {\sqrt n}{nh} = \frac {1}{h\sqrt n} [/tex]

[tex] a_n \leq \frac {1}{nh^2} [/tex] and [tex] a_n \rightarrow 0 [/tex]

Now my question is: Suppose you are asked to find numbers [tex] N_1, N_2, N_3 [/tex] such that

(a) [tex]\frac {n}{2^n} < \frac {1}{10}[/tex] for every [tex] n > N_1 [/tex]

(b) [tex]\frac {n}{2^n} < \frac {1}{100}[/tex] for every [tex] n > N_2 [/tex]

(c) [tex]\frac {n}{2^n} < \frac {1}{1000}[/tex] for every [tex] n > N_3 [/tex]

Do I have to guess and check for these numbers or can I somehow use the proof of this sequence? Because what if [tex] \epsilon [/tex] becomes smaller and smaller? Then there must be some method to find [tex] N_1, N_2, N_3 [/tex]

Thanks a lot for any help or advice

EDIT: Its [tex] \frac {n}{2^n} < \frac {1}{10} [/tex] for every [tex] n > N_1 [/tex] and the same with the others

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# Limits of Sequences

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