Prove that: $\zeta(3)=\lim_{N\to \infty}{1\over N}\sum_{k=1}^{N}{1\over k^{\phi^2}\ln{\left(1+{k^{\phi^{-2}}\over N}\right)}}$ 
Show that:
  $$\zeta(3)=\lim_{N\to \infty}{1\over N}\sum_{k=1}^{N}{1\over k^{\phi^2}\ln{\left(1+{k^{\phi^{-2}}\over N}\right)}}$$
  where $\phi$ is the golden ratio and $\zeta(3)$ is the Apery constant.

My try:
It converges very slowly when I tested it a sum calculator.
Let summed out
$$\zeta(3)=\lim_{N\to\infty}{1\over \ln{\left(1+{1\over N}\right)}}+{1\over 2^{\phi^{2}}\ln{\left(1+{2^{\phi^{-2}}\over N}\right)}}+{1\over 3^{\phi^{2}}\ln{\left(1+{3^{\phi^{-2}}\over N}\right)}}\cdots\tag1$$
$$\zeta(3)=1+{1\over2^3}+{1\over3^3}+\cdots$$
$$\zeta(\phi^2)=1+{1\over2^{\phi^2}}+{1\over3^{\phi^2}}+\cdots$$
I wonder is there a sum of this form
$$S={1\over \ln{A_1}}+{1\over \ln{A_2}^2}+{1\over \ln{A_3}^3}+\cdots$$
May be this sum could be useful
$$\ln{\left(n^s\over n^s-1\right)}=\sum_{r=1}^{\infty}{1\over n^{sr}\cdot{r}}$$
I wonder if we could apply geometric sum[sum to $\infty$] to $(1)$
$$S_{\infty}={a\over 1-r}$$
$$=1+{1\over r}+{1\over r^2}+{1\over r^3}+\cdots$$
I need help on how prove (1)
 A: Rough estimate: We have $\log(1+x) \approx x$ as $|x| \ll 1$ so for large $N$ the sum is approximately
$$ \frac{1}{N}\sum_{k=1}^N \frac{1}{k^{\phi^2}\log\left(1 + \frac{k^{\phi^{-2}}}{N}\right)} \approx \sum_{k=1}^N \frac{1}{k^{\phi^2 +\phi^{-2}}} = \sum_{k=1}^N \frac{1}{k^3}$$
since $\phi^2 + \phi^{-2} = 3$. To prove this more rigorously we need to control the error in this approximation.

Proof: Let $S_N = \frac{1}{N}\sum_{k=1}^N \frac{1}{k^{\phi^2}\log\left(1 + \frac{k^{\phi^{-2}}}{N}\right)}$ then
$$S_N - \sum_{k=1}^N\frac{1}{k^3} = \sum_{k=1}^N\frac{1}{k^3}\left[\frac{\frac{k^{\phi^{-2}}}{N}-\log\left(1 + \frac{k^{\phi^{-2}}}{N}\right)}{\log\left(1 + \frac{k^{\phi^{-2}}}{N}\right)}\right]$$
Now$^{(1)}$ $0 \leq \frac{x - \log(1+x)}{\log(1+x)} \leq \frac{x}{2}$ for $x\geq 0$ giving us
$$0 \leq S_N - \sum_{k=1}^N\frac{1}{k^3} \leq \frac{1}{2N}\sum_{k=1}^N \frac{1}{k^{3 - \phi^{-2}}}$$
which converges to $0$ as $N\to\infty$ as the series on the right hand side is convergent ($3-\phi^{-2} \simeq 2.6 > 1$).

Generalization: A generalization of the problem which can be proven in exactly the same way is: if $0<A$, $0<B<1$ with $A+B>1$ then
$$\zeta(A+B) = \lim_{N\to\infty}\frac{1}{N}\sum_{k=1}^N \frac{1}{k^{A}\log\left(1 + \frac{k^{B}}{N}\right)}$$
$^{(1)}$ Consider $f(x) = \frac{x}{2} - \frac{x}{\log(1+x)} +1$ then $\lim_{x\to 0}f(x) = 0$ and $f'(x) = \frac{1}{2} - \frac{\frac{x}{1+x} - \log(1+x)}{\log^2(1+x)}$ $= \frac{x}{6} - \frac{x^2}{8} + \ldots > 0$ for small enough $x$ (but turns out this holds for all $x>0$).
