1-D Simple Harmonic Oscillator with Dirac Notation I'm having trouble understanding some of the subtleties of working with bras and kets when considering the standard 1-D quantum oscillator.
Say I am given a state vector at $t=0$, $$|\Psi(t=0)\rangle = A \sum_{q=0}^{Q_o} \frac{1}{q+i}|\phi_q\rangle$$
Where $Q_o$ is a positive finite integer.
I am given the number operator $$\hat{N}|\phi_q\rangle = q|\phi_q\rangle$$ where $\hat{N} = \hat{a}^\dagger\hat{a}$, and the energy eigenvalues of the quantum oscillator are known as $$E_q = (q+\frac{1}{2})\hbar\omega_o$$ for $q = 0,1,2,3....$
Now, I am tasked to answer the following questions regarding the state vector $|\Psi(t=0)\rangle$:

[1]
Find an equation for $A$ which normalizes the state vector $|\Psi(t=0)\rangle$
In wave mechanics, the coefficient for normalization is typically found by enforcing that $\int|\Psi|^2 dx = 1$ for all space, and then solving for the coefficient within $\Psi$.
If I am to let $Q_o = 100$, $$|\Psi(t=0)\rangle = |\Psi(0) \rangle = A \sum_{q=0}^{100} \frac{1}{q+i}|\phi_q\rangle$$
I know that a normalized state vector will follow $$\langle \Psi(t)|\Psi(t)\rangle = 1$$
So if I have $$\langle \Psi(0)| = (|\Psi(0)\rangle)^* = A^*\sum_{q=0}^{100} \frac{1}{q-i}\langle\phi_q|$$
then, $$\langle \Psi(0)|\Psi(0)\rangle = (A^*\sum_{q=0}^{100} \frac{1}{q-i}\langle\phi_q|)(A \sum_{q=0}^{100} \frac{1}{q+i}|\phi_q\rangle)=1$$
Now if $A = A^*$, am I able to state that $$A^2\sum_{q=0}^{100}\sum_{q=0}^{100}\frac{1}{q+i}\frac{1}{q-i}\langle\phi_q|\phi_q\rangle = 1$$
where $$\langle\phi_q|\phi_q\rangle = \langle q|\hat{N} |q\rangle = q\langle q|q\rangle = q$$ (as $\langle q|q\rangle = 1$ due to orthonormality), and then solve for $A$? 
Or am I horribly mistaken in terms of how bras and kets work?
From what I have so far, I would obtain that $$A^2\sum_{q=0}^{100}\sum_{q=0}^{100}\frac{q}{(q+i)(q-i)} = 1$$ and $$A = \sqrt{\sum_{q=0}^{100}\sum_{q=0}^{100}\frac{(q+i)(q-i)}{q}}$$
Hopefully someone will be able to clarify my misunderstandings here..

[2]
Evaluate $\langle \phi_q|\hat{N}$
Am I able to simply state here that $$\langle \phi_q|\hat{N} = \sum_{q}N_{p,q}\langle \phi_q| = \sum_{q}N_{p,q}\langle q|\hat{a}^\dagger$$
where $N_{p,q}$ are the matrix elements of the number operator $\hat{N}$?
I don't quite know what I am supposed to end up with as an answer or proceed further to work with the sum produced, provided it is even correct.

[3]
What is the state vector for times $t \geq 0$, and the expectation value for $\hat{a}^\dagger$ for these times?
From my understanding, I can simply tack on time dependence to the original state vector. For example, $$|\Psi(t)\rangle = |\phi_q\rangle e^{\frac{-iE_qt}{\hbar}}$$
then, will the expectation value $\hat{a}^\dagger$ just be $$\langle \Psi(t)|\hat{a}^\dagger|\Psi(t)\rangle$$
I'm not sure how to actually calculate this expectation value, but looking at my notes, I see that $$\langle p|\hat{a}^\dagger|q\rangle = \sqrt{q+1}\delta_{p,q+1}$$
Does this imply that I can write $$\langle \hat{a}^\dagger\rangle = \langle \phi_q e^{-i\omega_o t}|\hat{a}^\dagger|\phi_q e^{-i\omega_o t}\rangle = \sqrt{q+1}e^{-i\omega_o t}$$?

Sorry for the length of the question, but hopefully it will make it easy to figure out where I'm going wrong in my approach of these problems. Thanks.
 A: 
Number 1. In general terms.

A simple harmonic operator is in a state (at t = 0), with state vector
$\Psi (x,  0) = A \sum_{n=0}^{\infty} c^n|\phi_n (x)\rangle$
The normalisation constant A, up to a constant phase, is:
$$A^{-2} =  \sum_{n=0}^{\infty} |c|^{2n} =\frac {1}{1-|c|^2}$$
Leads to $$A  = \sqrt {1-|c|^2}$$
The  time evolved state vector is
$$\Psi (x,  t) = A e^{-i\omega t/2}  \sum_{n=0}^{\infty} c^ne^{-in\omega t}|\phi_n (x)\rangle $$

Number 3.

$${\displaystyle {\begin{aligned}{\hat {x}}&={\sqrt {{\frac {\hbar }{2}}{\frac {1}{m\omega }}}}(a^{\dagger }+a)\\{\hat {p}}&=i{\sqrt {{\frac {\hbar }{2}}m\omega }}(a^{\dagger }-a)~.\end{aligned}}}$$

This means that a acts on |n⟩ to produce, up to a multiplicative constant, |n–1⟩, and a† acts on |n⟩ to produce |n+1⟩. For this reason, a is called a "lowering operator", and a† a "raising operator". The two operators together are called ladder operators. In quantum field theory, a and a† are alternatively called "annihilation" and "creation" operators because they destroy and create particles, which correspond to our quanta of energy.
Given any energy eigenstate, we can act on it with the lowering operator, a, to produce another eigenstate with ħω less energy. By repeated application of the lowering operator, it seems that we can produce energy eigenstates down to E = −∞. However, since

$$ {\displaystyle n=\langle n\mid N\mid n\rangle =\langle n\mid a^{\dagger }a\mid n\rangle =\left(a\mid n\rangle \right)^{\dagger }a\mid n\rangle \geqslant 0}$$
 Wikipedia Ladder Operators
