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 $\sum\_{k=1}^\infty f\_k < \infty$.



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> 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.
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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.



Does this infinite sum arising from separation of variables converge?
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 $[0,2R]$.


