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3

Multiply numerator and denominator by $\frac{2a}{\log x}$: $$\frac{2a}{1-2a\frac{\log\log x}{\log x}}.$$ $a<1/2$ and $\log(x) < x$; so $2a\log\log x/\log x < 1$, and $$f(x)=2a\sum_{n=0}^\infty \left(2a\frac{\log\log x}{\log x}\right)^n.$$ So, $f(x)\sim 2a$.

2

Given $$f(x) = \frac{ 2 x - 2 }{ x^2 - 2 x + 4 }. \tag 1$$ Assuming you want to expand $f(x)$. Let $$\phi_\pm = 1 \pm \mathtt{i} \sqrt{3}. \tag 2$$ We can write (1) as $$f(x) = \frac{ 1 }{ x - \phi_+ } + \frac{ 1 }{ x - \phi_- }. \tag 3$$ Whence f(x) = - \sum_{k=0}^\infty \left( \frac{1}{{\phi_+}^{k+1}} + ... 2 The given f(x) is given by \begin{align} f(x) = \frac{2(x-1)}{x^2 - 2x + 4} \end{align} and can be seen as, where a = 1 + \sqrt{3} i and b = 1 - \sqrt{3} i, such that ab=4, \begin{align} f(x) = \frac{-2 \, (1 - x)}{ab \, \left(1 - \frac{x}{a}\right) \left( 1 - \frac{x}{b}\right)} \end{align} for which \begin{align} \ln f(x) &= \ln\left(- ... 2 You can find a formula on Wikipedia: \sum \limits _{n=1} ^\infty H_n ^{(m)} x^n = \frac {\mathrm{Li}_m (z)} {1-z}, where \mathrm{Li}_m is the polylogarithm. I doubt that you will be able to find something involving more elementary functions. 1 Let m=\lceil -\mathrm ep\rceil. Then the ratio of magnitudes of successive terms beyond m is less than 1/\mathrm e. Now choose n such that \frac{|p|^m}{m!}\frac1{1-1/\mathrm e}\mathrm e^{-(n-m)}\le\frac12\mathrm e^p\;.  Since the $m$-th term has magnitude $|p|^m/m!$ and the magnitudes of the remaining terms are bounded by a geometric sequence ...

1

Yes, your method is correct. Basically, to find a power series solution, we would assume the existence of $\{a_n\}$, such that $y(x) = \sum a_n x^n$. The next thing to do would be finding a recurrent form of $a_n$, which leads to finding a closed form of it (in case it exists). What you have done is that throughout your method, you found the expression of ...

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