Let $\mathfrak h$ be the $1$-particle Hilbert space, $|\psi_i\rangle$ an orthonormal basis and $\mathcal F$ the bosonic Fock space (edit : for indistinguishable bosons), which is given by :
$$\mathcal F = \bigoplus_{n=0}^{\infty}S^n\mathfrak h$$
Here, $S^n \mathfrak h$ is the symmetrized $n$-fold tensor product of $\mathfrak h$, ie the space of $n$-particle states.
$\mathcal F$ is a Hilbert space, with the different $n$-particle subspaces being orthogonal. Therefore, there are $n$-particle states with arbitrary high values of $n$ as well as states which are superposition involving arbitrary high values of $n$.
Now, let $P_n$ be the orthogonal projection onto the $n$-particle subspace, $\underline \alpha = \{\alpha_i\}$ a collection of complex numbers and set (formally, for now) :
$$|\underline \alpha\hspace{0.3mm}\rangle = \exp\bigg\{-\frac{1}{2}\sum_i|\alpha_i|^2\bigg\}\exp\bigg\{-\sum_i\alpha_ia_i^{\dagger}\bigg\}\exp\bigg\{\sum_i\alpha^*_ia_i\bigg\}|0\rangle$$
Then, we have :
\begin{align}
P_n |\underline \alpha\hspace{0.3mm}\rangle &= \frac{1}{n!}\exp\bigg\{-\frac{1}{2}\sum_i|\alpha_i|^2\bigg\}\bigg(-\sum_i\alpha_ia_i^{\dagger}\bigg)^n|0\rangle
\end{align}
In particular, we have :
\begin{align}
P_1|\underline \alpha\hspace{0.3mm}\rangle &= \exp\bigg\{-\frac{1}{2}\sum_i|\alpha_i|^2\bigg\}\bigg(-\sum_i\alpha_ia_i^{\dagger}\bigg)|0\rangle\\
&= -\exp\bigg\{-\frac{1}{2}\sum_i|\alpha_i|^2\bigg\}\sum_i \alpha_i|\psi_i\rangle
\end{align}
This is a well-defined state of $\mathfrak h$ only if $\sum_i |\alpha_i|^2 <\infty$, ie this condition is necessary. If this condition holds, we can define :
\begin{align}
\sqrt{\sum_i |\alpha_i|^2} &=\sqrt{\sum_i |\alpha_i|^2} \\
b &= \frac{1}{\sqrt{\sum_i |\alpha_i|^2}}\sum_i \alpha_i^* a_i
\end{align}
and check that $[b,b^\dagger] = 1$ : $b$ is an annihilation operator and :
$$|\underline \alpha\hspace{0.3mm}\rangle = \exp\left(-\frac{1}{2}|\underline\alpha|^2\right)\exp(-|\underline\alpha|b^\dagger)\exp(|\underline\alpha|b)|0\rangle
$$
We can compute the norm of this state by projecting it onto every $n$-particle subspace :
\begin{align}
\langle \hspace{.3mm} \underline\alpha|\underline \alpha\hspace{0.3mm}\rangle &= e^{-|\underline\alpha|^2} \frac{1}{n!^2}\sum_{n=0}^\infty\langle 0 |e^{-|\underline\alpha|b}P_ne^{-|\underline\alpha|b^\dagger}|0\rangle \\
&= e^{-|\underline\alpha|^2} \sum_{n=0}^\infty\frac{|\underline\alpha|^{2n}}{n!^2}\langle 0 |b^n (b^\dagger)^n|0\rangle \\
&=e^{-|\underline\alpha|^2} \sum_{n=0}^\infty\frac{|\underline\alpha|^{2n}}{n!} \\
&= 1
\end{align}
Edit :
The Fock space is generated by states with a definite and finite number of particles. But, given an orthonormal basis of such states, you can form linear combination with $L^2$ coefficients to form other states in the Fock space. In general, a (normalized) state $|\Psi\rangle\in\mathcal F$ has :
$$1 = \langle \Psi|\Psi\rangle = \sum_{n=0}^{+\infty} \langle\Psi|P_n|\Psi\rangle$$
This state has a probability $p_n = \langle\Psi|P_n|\Psi\rangle$ to contain $n$-particles. This probability distribution does not need to have finite support and it is even possible for it to have an infinite expectation value : $\sum_{n=0}^{+\infty} np_n = +\infty$. However, it must be a probability distribution, ie it must sum up to $1$.
Edit 2 : As mentionned in the comments, given an orthonormal basis of $\mathfrak h$, we can build an orthonormal basis of the bosonic Fock space using occupation numbers, with vectors of the forms $|\{n_i\}\rangle$ where the $n_i$ are non-negative integers, all but a finite number of whom are zero.
If we consider the subset of those vectors with $\sum_i n_i = n$, we get an orthonormal basis for the $n$-particle subspace. By concatening those, we get a basis for the full Fock space.
An $n$-particle subspace (with $n$ fixed) can be finite dimensional if $\dim\mathfrak h<\infty$, but the bosonic Fock space is always infinite dimensional.
