## 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: $$$$a_0\left(\lambda(1-\gamma-\ln \lambda)-1+\gamma\right)+\displaystyle\sum_{n=1}^{\infty}a_n\left(\frac{1}{n}(\gamma-1)+\frac{\zeta(n+1)}{n+1}(1-\lambda^{n+1})\right)=0\quad…(1)$$$$

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.