TITLE: Does this infinite sum arising from separation of variables converge?
QUESTION [5 upvotes]: This problem came up in a PDE where I used separation of variables to formally get a solution. Now I need to know whether that formal solution is sensible.
Let $a_k >0$ be an increasing sequence of real numbers. Let $f_k$ be real numbers.
Let $R \in (0,\infty)$ and $y \in (0,\infty)$. Let 
$$v(y) = \sum_{k=1}^\infty f_k\left(B_1(k,R)e^{-a_ky} -B_2(k,R)e^{a_ky}\right)$$
where 
$$B_1(k,R) = \frac{e^{a_kR}}{e^{a_kR}-e^{-a_kR}}\quad\text{and}\quad B_2(k,R) = \frac{e^{-a_kR}}{e^{a_kR}-e^{-a_kR}}.$$
We also know $\sum_{k=1}^\infty f_k < \infty$.

Is $v(y)$ finite for all $y$ and $R$ (i.e. does the infinite sum exist)? Also I would like to know whether the series is uniformly convergent (so whether the partial sums uniformly converge) so that I can do term by term integration.

I don't see why the sum even exists...
Remark: In fact, $a_k^2$ are the eigenvalues of the Neumann Laplacian, and $f_k := (u,\varphi_k)_{L^2}\varphi_k(x)$ on some bounded domain $\Omega$ where $\varphi_k$ are the eigenvectors of the Neumann Laplacian.

REPLY [4 votes]: For fixed $R > 0$, $B_1(k,R) \sim 1$ as $k$ goes to infinity, while $B_2(k,R) \sim e^{- 2 a_k R}$. The bound on $B_1$ combined with the summability of the $f_k$ takes care of one half of your sum.
The other half has a general term of order $f_k e^{-2a_kR + a_ky}$. If the $f_k$ are only generic summable reals as your post seems to imply, then your sum is convergent only for $y \leq 2R$, divergent otherwise. But hopefully, you are working on some bounded domain such that $y$ never exceeds $2R$, aren't you ?
Also, uniform (in $y$) convergence holds on $[0,2R]$.