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Arrange the following functions in increasing asymptotic order:

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    This has nothing to do with the divisor-counting-function which is the only tag you've applied to the question. None of the content of the question has anything to do with power series either which is what you have as the title. Please use a more accurate and descriptive title and use appropriate tags when asking questions. The more effort you put into asking a question, the more effort we will put into answering. As it stands, it appears you have put next to no effort in and so that is likely what you will receive in return. – 2017-01-13
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    As for the content of your actual question "why does option $e$ get ranked as #4 in the list" what is it that you don't understand about that? All of $(a),(c),(d)$ are polynomial (*note that $\log^9 n < \sqrt{n}$ for large enough $n$ so $n\log^9n$(e)$ is exponential. $n^\alpha \lll \beta^n$ for all $\alpha$ and all $\beta>1$. Then the only thing left to notice is that $e>1.0000001$ – 2017-01-13
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    according to u what is right answer? – 2017-01-13
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    not getting i use my calculator and calculate its value which still remain 1.0000.... for large n – 2017-01-13
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    "which still remain 1.0000... **for large n**" but apparently not large enough $n$. If you are using numbers that you can conceive, then you aren't thinking big enough. $1.000001^{10^{10^{10^{100}}}}$ is most certainly much bigger than anything you would normally think of and is many more than simply trillions of digits long. Don't trust technology too much. – 2017-01-13
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    @sittian: You're not checking for large enough $n$ then. Try $n = 10^{30}$. At this value, e^n > 1.0000001^n > n^{7/4} > nlog(n)^9 > n^{1/3}. In general though, plugging in large numbers is not a very reliable method, as there's no way to know how big it needs to be without analytically finding it, and you may overflow your calculator before even reaching that number. It should follow fairly easily that $e^n > 1.0000001^n > n^{7/4} > n^{1/3}$, so you just have to determine the correct location of c. Which JMoravitz did for you already. – 2017-01-13
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    k it lead me in imaginary situation. got it . tie . cn nyone remove downvote else i will not able to ask another question . – 2017-01-13

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In addition to JMoravitz's comment, note that:

$$\lim \limits_{n \to infinity} \frac{1.0000001^n}{n^a}$$ is infinite. So we can apply L'Hopital's Rule. The derivative of $$1.0000001^n = 1.0000001^n ln(1.0000001)$$ and the derivative of $$n^a = an^{a-1}$$ which is still infinity in the limit that $n$ goes to infinity. The $a^{th} $ iteration of L'Hopital's rule would yield: $$\lim \limits_{n \to infinity} \frac{1.0000001^n ln(1.0000001)^a}{a!}$$ which is still infinity. Therefore, $1.0000001^n$ is asymptotically larger than any finite degree polynomial, despite its appearance.