Prove
$$\lim_{n\ \mapsto\ \infty}4^n \psi^{(2)}(2^n)-\psi^{(2)}(1)=2\zeta(3)-1\tag{1}$$
where $\psi^{(m)}(x)$ is the polygamma function and $\zeta$ is the Riemann zeta function.
This problem was proposed by a friend and solved using the series expansion of digamma function:
$$\psi(x)=\ln x+O\left(\frac1x\right)$$ and by differentiating both sides $k$ times we get
$$\psi^{(k)}(x)=\frac{(-1)^{k-1}(k-1)!}{x^k}+O\left(\frac1{x^{k+1}}\right)$$
and by setting $k=2$ and $x=2^n$ then letting $n\mapsto \infty$, the result follows.
The question here is can we prove (1) in a different way?
UPDATE:: You can find the generalization of this problem in the book, Almost impossible integrals, sums and series page 90.
Solution 1. Differentiating the partial fraction decomposition of the digamma function
$$ \psi(z) = -\gamma + \sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+z} \right) $$
twice, we get
$$ \psi^{(2)}(z) = -\sum_{n=0}^{\infty} \frac{2}{(n+z)^3}. $$
From this, we easily find that $\psi^{(2)}(1) = -2\zeta(3)$. Moreover, for $z > 1$ we have
$$ \frac{1}{2z^2} = \int_{0}^{\infty} \frac{1}{(x+z)^3} \, \mathrm{d}x \leq \sum_{n=0}^{\infty} \frac{1}{(n+z)^3} \leq \int_{0}^{\infty} \frac{1}{(x+z-1)^3} \, \mathrm{d}x = \frac{1}{2(z-1)^2},$$
and so, we deduce that $\lim_{z\to\infty} z^2\psi^{(2)}(z) = -1$. Combining these two computations prove the desired result. This solution is essentially unwinding the derivation of the asymptotic form $\psi(z) = \log z + \mathcal{O}(1/z)$, thus is not so different from OP's solution.
Solution 2. Alternatively, starting from the identity
$$ \psi(z) = -\gamma + \int_{0}^{\infty} \frac{1-e^{-(z-1)s}}{e^s-1} \mathrm{d}s, $$
we get
$$ \psi^{(2)}(z) = - \int_{0}^{\infty} \frac{s^2 e^{-(z-1)s}}{e^s - 1} \, \mathrm{d}s, $$
and so, $\psi^{(2)}(1) = -\Gamma(3)\zeta(3) = -2\zeta(3)$ and
$$ z^2 \psi^{(2)}(z) \stackrel{(t=zs)}= - \int_{0}^{\infty} \frac{t/z}{1 - e^{-t/z}} \, t e^{-t} \, \mathrm{d}t \xrightarrow[z\to\infty]{} - \int_{0}^{\infty} t e^{-t} \, \mathrm{d}t = -1 $$
by the dominated convergence theorem. This completes the proof.
Solution 3. Here we assume that we already know $\psi^{(2)}(1) = -2\zeta(3)$. By log-differentiating the Legendre duplication formula, we get
$$ \psi(z) = \log 2 + \frac{1}{2} \left[ \psi\Big(\frac{z}{2}\Big) + \psi\Big(\frac{z+1}{2}\Big) \right]. $$
Then the repeated application of this identity yields
$$ \psi(z) = n \log 2 + \frac{1}{2^n} \sum_{k=0}^{2^n-1} \psi\Big(\frac{z+k}{2^n}\Big). $$
Differentiating both sides twice followed by plugging $z = 2^n$ and multiplying $4^n$ to both sides, we get
$$ 4^n \psi^{(2)}(2^n) = \frac{1}{2^n} \sum_{k=0}^{2^n-1} \psi^{(2)}\Big(1+\frac{k}{2^n}\Big). $$
As $n\to\infty$, this converges to
$$ \lim_{n\to\infty} 4^n \psi^{(2)}(2^n) = \int_{0}^{1} \psi^{(2)}(1+x) \mathrm{d}x = \psi^{(1)}(2) - \psi^{(1)}(1) = -1. $$