How to show this infinite sum converges uniformly? Let $f_k$ be a real numbers such that $\sum_{k=1}^\infty f_k < \infty$. For each $R > 0$, define the convergent sum
$$v(R) = \sum_{k=1}^\infty f_k(b_k(R)e^{-ky} - c_k(R)e^{ky})$$
where $0 \leq y \leq R$ and where $b_k$ and $c_k$ are functions of $k$ and $R$ and satisfy $$b_k(R) = 1+ \frac{1}{e^{2kR}-1}\quad \text{and}\quad c_k(R) = b_k(R)-1 = \frac{1}{e^{2kR}-1}$$ 

Apparently this sum uniformly converges (see this) in $n$ as function of $R$. How do I show this? I.e. if $v_n(R) = \sum_{k=1}^n f_k(b_k(R)e^{-ky} - c_k(R)e^{ky})$ how do I show that $v_n \to v$ uniformly as function of $R$?

I didn't understand how the explanation is given in that thread and I think my question is too basic to be asked there. Both the coefficients have look "like" $e^{-2kR}$ so it seems like it should be uniform but I don't know the argument.

Assume that $\sum_k f_k$ converges absolutely.
Note that 
$$\begin{align}
|b_k(R)e^{-ky}-c_k(R)e^{ky}|&=\left|\frac{e^{2kR}e^{-ky}-e^{ky}}{e^{2kR}-1}\right|\\\\
&=\left|\frac{\sinh k(R-y)}{\sinh kR}\right|\\\\
&<1
\end{align}$$
Thus
$$\begin{align}
\left|v(R)   \right|&=\left|\sum_{k=1}^{\infty}f_k\left(b_k(R)e^{-ky}-c_k(R)e^{ky}\right)   \right|\\\\
&\le \sum_{k=1}^{\infty}|f_k|\,\left|\left(b_k(R)e^{-ky}-c_k(R)e^{ky}\right)\right| \\\\
&\le  \sum_{k=1}^{\infty}|f_k|  \\\\
\end{align}$$
which converges by hypothesis!

Assume that $\sum_k |f_k|^2$ converges.
Then,
$$\begin{align}
\left|v(R)   \right|^2&=\left|\sum_{k=1}^{\infty}f_k\left(b_k(R)e^{-ky}-c_k(R)e^{ky}\right)   \right|^2\\\\
&\le \sum_{k=1}^{\infty}|f_k|^2\,\,\sum_{k=1}^{\infty}\left|\left(b_k(R)e^{-ky}-c_k(R)e^{ky}\right)\right|^2 \\\\
&=\sum_{k=1}^{\infty}|f_k|^2\,\,\sum_{k=1}^{\infty}\left|\frac{\sinh k(R-y)}{\sinh kR}\right|^2
\end{align}$$
The first term converges by hypothesis.  For the second term, note that 
$$\left|\frac{\sinh k(R-y)}{\sinh kR}\right|^2\le e^{-2ky}$$
and $\sum_{k=1}^{\infty} e^{-2ky}$ converges for any fixed $y>0$.
