Solve for $a_n’$s, $\sum_{n=1}^{\infty}a_n\left(\frac{1}{n}(\gamma-1)+\frac{\zeta(n+1)}{n+1}(1-\lambda^{n+1})\right)=0$

I am looking for a non-trivial uniformly convergent series $ \displaystyle\sum_{n=0}^{\infty}a_nx^n$ on $ [0,1]$ , where each $ a_n$ is independent of $ \lambda,0<\lambda\le 1$ and satisfying: \begin{equation}a_0\left(\lambda(1-\gamma-\ln \lambda)-1+\gamma\right)+\displaystyle\sum_{n=1}^{\infty}a_n\left(\frac{1}{n}(\gamma-1)+\frac{\zeta(n+1)}{n+1}(1-\lambda^{n+1})\right)=0\quad…(1) \end{equation}

Here $ \gamma$ is Euler-Mascheroni constant and $ \zeta$ is Riemann zeta function. I don’t know if such a series would exist. We have the following result on $ [0,1]$ :

If $ \sum_{n=0}^\infty|a_n|<\infty$ , then $ \sum_{n=0}^\infty a_nx^n$ converges uniformly for $ x\in[0,1]$ .

So we need to find the sequence $ \{a_n\}$ with $ \sum_{n=0}^\infty|a_n|<\infty$ satisfying equation $ (1)$ .

Setting $ a_0=0$ might simplify our question.