For any unbounded sequence $(b_k)$ and any sequence $(\lambda_n)$, you can find a sequence $(a_k)$ such that for any $n$, $\sum_{k=1}^\infty a_k b_k^n$ converges absolutely to $\lambda_n$ :
Suppose you have constructed $a_k$ up to some index $k_0$, such that for all $m < n$, $\sum_{k=1}^{k_0} a_k b_k^m = \lambda_m$. The goal now is to extend this up to some index $k_1$ such that for all $m \le n$, $\sum_{k=1}^{k_1} a_k b_k^m = \lambda_m$, with "arbitrarily small coefficients".
This means we have to add $y = \lambda_n - \sum_{k=1}^{k_0} a_k b_k^n$ to the partial sum of the $n$th series, and $0$ to the first $n-1$ series. In order to do that, look at the next $n-1$ distincts values of $b_k$ (for $k > k_0$), say they are $c_1,c_2, \ldots c_{n-1}$, look at the Vandermonde matrix
$ M(X) = \begin{pmatrix} c_1^1 & \ldots & c_{n-1}^1 & X^1 \\ \vdots & & \vdots & \vdots \\ c_1^n & \ldots & c_{n-1}^n & X^n \end{pmatrix}$
and solve for the vector $A(X)$ in the equation $M(X) A(X) = (0,0,\ldots,0,y)$. We get that $A(X)$ has to be $y$ times the $n$th column of $M(X)^{-1}$. Thus $A(X)$ is $y/det(M(X))$ times the transpose of the $n$th row of the comatrix of $M(X)$. If you look carefully at this row, each entry there is a polynomial in $X$ of degree $n-1$, except the last one which is a constant. Furthermore, the determinant of $M(X)$ is of degree $n$, so the entries of the vector $A(X)$ are rational fractions of degree $-1$ and $-n$.
In particular, for $X$ large enough, $\sum_{1 \le i,j \le n-1} |A_i(X)c_i^j| < 2^{-n}$ Thus, if you pick $k_1$ such that $b_{k_1}$ is large enough, you will get from $A(b_{k_1})$ the coefficients to put in front of $c_1^n, c_2^n, \ldots, c_{n-1}^n, b_{k_1}^n$ such that all the partial sums are what we want, and the absolute contribution to what we added to each partial sum is less than $2^{-n-1}$.
Now, repeat this procedure for all $n$, and you get a suitable sequence $(a_k)$.