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.

I'm just fumbling here, but if you compute a bit you get
$$
 b_k(R) e^{-ky} - c_k(R) e^{ky}
 =
 \frac 1 {e^{2kR}-1} \left( e^{2kR} e^{-ky} - e^{ky} \right).
$$
For $R$ large enough, you can easily bound this by some constant times
$$
e^{-ky} - e^{k(y-2R)}.
$$
Since $R \ge y$, the second term will be $\le e^{-ky}$. So for large $R$, you can bound your summand by
$$
  | f_k (b_k(R) e^{-ky} - c_k(R) e^{ky}) | \le C e^{-ky} |f_k| \le C |f_k|.
$$
Since the $\sum f_k$ converges, so does this new sum (assuming $f_k\ge 0$), and since the bound doesn't depend on $R$, it converges uniformly in $R$ (again, for large $R$).
Please double check all of this, I'm not sure if I messed up along the way...
EDIT: Ouch, Dr. MV had the more elegant idea obviously.
