Approximating a series for asymptotic upper bound So I stumbled upon a question here to find an asymptotic bound for the recurrence
$$
T(n) = 4T(n/4) + \left( \frac {n} {\log n} \right)^2 $$

The solution shows a solution using master method to prove the answer as $O(n)$
However, I tried to solve it using substitution as:

$$T(\frac {n} {4^i}) = 4^{i+1} T(\frac {n} {4^{i+1}} ) + \sum_{j=0}^{i} \frac {n^2} {4^j {(\log \frac{n}{4^j})^2} }$$

I then get
$$
T(n) = 4^{x+1}(Base Constant) + \sum_{j=0}^{logn -1} \frac {n^2} {4^j {(log \frac{n}{4^j})^2} }$$
$x=\log n$

Now for the summation:
$$n^2\sum_{j=0}^{\log n -1} \frac {1} {4^j {(log(n) - j\log (4))^2} }$$

All logarithms to be taken base 4:
$n^2\sum_{j=0}^{\log n -1} \frac {1} {4^j {(\log (n) - j)^2} }$

Now, the first substitution I made was $t = \log n -j $

$n^2\sum_{j=1}^{\log n -1} \frac {4^t} {4^{\log n} {t^2} }$

= $n^2\sum_{j=1}^{\log n -1} \frac {4^t} {n {t^2} }$
= $n\sum_{j=1}^{\log n -1} \frac {4^t} { {t^2} }$

Using wolfram and wkipedia, the sum comes out to $Li2(4) -n Lerch Function(4,2,\log (n)-1)$

I could simple assume this is constant and that would prove my bound to be $O(n)$ as the master theore suggests, but I ahve no proof to bound the Lerch and dilogarithmic function.

Another substitution I tried was

$$ t=4^j (\log n -j)^2 $$ 
which gives me
$$n^2 H(n) - H((\log n)^2) $$ where $H(n)$ is the nth harmonic number
I again could not reach a bound for this, besides $\log (n) - \log (\log n) $
which is quite different from my first substitution.

If someone could help me with what error I'm making, or why the substitutions yield different results that would be great.
To make things a bit simpler, let us assume that we use base 4 logarithm. Following your steps, for $n = 4^k$ and $T(1) = 0$ we get
$$
\begin{align*}
T(n) &amp;= \frac{n^2}{(\log_4 n)^2} + 4\frac{(n/4)^2}{(\log_4 n-1)^2} + 16\frac{(n/16)^2}{(\log_4 n - 2)^2} + \cdots \\ &amp;=
\frac{n^2}{(\log_4 n)^2} \left[ 1 + \frac{1}{4(1-1/\log_4 n)^2} + \frac{1}{16(1-2/\log_4 n)^2} + \cdots \right].
\end{align*}
$$
Let us now use the series expansion
$$
\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + \cdots.
$$
Substituting this in the expression above, we get
$$
\begin{align*}
\frac{T(n)}{n^2/\log^2_4n} &amp;= 1 + \frac{1}{4} + \frac{1}{16} + \cdots \\ &amp;+
2\left[ \frac{1}{4\log_4 n} + \frac{2}{16\log_4 n} + \cdots \right] \\ &amp;+
3 \left[ \frac{1}{4\log_4^2n} + \frac{2^2}{16\log_4^2n} + \cdots \right] + \cdots \\ &amp;\approx \frac{4}{3} + \frac{8/9}{\log_4 n} + \frac{20/9}{\log_4^2 n} + \cdots
\end{align*}
$$
The error in the approximation is polynomially small: for example 
$$
\sum_{k=0}^{\log_4 n-1} \frac{1}{4^k} = \frac{4}{3} - \frac{4/3}{4^{\log_4 n}} = \frac{4}{3} - \frac{4/3}{n}.
$$
The error is somewhat larger for the other summands, but still polynomially small. Taylor's theorem (or direct calculation) shows that if we truncate the series for $1/(1-x)^2$ after the term $(d+1)x^d$ then we get an error of $O(x^{d+1})$. Altogether, this shows that the following asymptotic statements hold:
$$
\begin{align*}
T(n) &amp;= \frac{4}{3} \frac{n^2}{\log_4^2 n} + O\left(\frac{n^2}{\log_4^3 n}\right), \\
T(n) &amp;= \frac{4}{3} \frac{n^2}{\log_4^2 n} + \frac{8}{9} \frac{n^2}{\log_4^3 n} + O\left(\frac{n^2}{\log_4^4 n}\right), \\
T(n) &amp;= \frac{4}{3} \frac{n^2}{\log_4^2 n} + \frac{8}{9} \frac{n^2}{\log_4^3 n} + \frac{20}{9} \frac{n^2}{\log_4^4 n} + O\left(\frac{n^2}{\log_4^5 n}\right), \ldots \\
\end{align*}
$$