Is there a nice way to express $\psi_1$ using this orthonormal sequence? Suppose that $H$ is a separable Hilbert space and $(\psi_n)_{n=1}^{\infty}$ is a complete orthonormal sequence in $H$. We define a sequence $(\phi_n)_{n=1}^{\infty}$ by
$$
\phi_n=\psi_1+\psi_{n+1}\ ;\  n=1,2,3,...
$$
Indeed, $(\phi_n)_{n=1}^{\infty}$ is a complete linearly independent sequence in $H$ since for any $f\in H$ such that $\langle f,\phi_n\rangle=0$ for all $n\in \Bbb N$, we have
$$\begin{align}
0 &= \langle f,\phi_n\rangle \\
&= \langle f,\psi_1 + \psi_n\rangle \\
&= \langle f,\psi_1\rangle+\langle f,\psi_n\rangle \\
\langle f,\psi_n\rangle &= -\langle f,\psi_1\rangle
\end{align}$$
and since $\langle f,\psi_n\rangle \to 0$ we must have $\langle f,\psi_1\rangle=0$. Consequently, $\langle f,\psi_n\rangle=0$ for all $n\in \Bbb N$ thus $f=0$ by the completeness of $(\psi_n)_{n=1}^{\infty}$. This shows that the linear span of $(\phi_n)_{n=1}^{\infty}$ is dense in $H$.
However, I am curious if there is a nice way to express $\psi_1$ as a series of $\phi_n$. Since each $\phi_n$ contains a part of $\psi_1$, clearly $(\phi_n)_{n=1}^{\infty}$ is not an orthogonal sequence and the representation of $\psi_1$ ought to be not unique. Right now, I know that there can't be the coefficients $c_n$ such that 
$$
\psi_1=\sum_{n=1}^{\infty}c_n\phi_n
$$
My reasoning is that I will be forced to choose $c_n$ such that $\sum c_n=1$ but that would give 
$$\begin{align}
\psi_1&=\sum_{n=1}^{\infty}c_n\phi_n \\
&=\sum_{n=1}^{\infty}c_n\psi_1+\sum_{n=1}^{\infty}c_n\psi_{n+1} \\
&= \psi_1+\sum_{n=2}^{\infty}c_{n-1}\psi_n \\
0&=\sum_{n=2}^{\infty}c_{n-1}\psi_n
\end{align}$$
which doesn't seems right either since it would mean that $c_1=c_2=...=0$, so such $c_n$'s don't exist.
Edit: It seems that the question is too confusing. By the phrase "nice form" I am not referring to the finite linear combination but rather anything in the spirit of expanding $f=\sum_{n=1}^{\infty}c_n\psi_n$. In this case we have $c_n=\langle f,\psi_n\rangle$ but this is possible because $(\psi_n)_{n=1}^{\infty}$ is orthogonal.

If $\phi_n= \psi_1 + \psi_{n+1}$, then with
$$x_n:=\sum_{k=1}^n \frac{1}{n} \phi_k =\sum_{k=1}^n\frac{\psi_1}{n}+\sum_{k=1}^n\frac{\psi_{k+1}}n= \psi_1 +\frac{1}{n}\sum^n_{k=1}\psi_{k+1}$$
you get
$$\langle x_n - \psi_1, x_n - \psi_1\rangle = 1+||x_n||^2 -2 \langle x_n ,\psi_1\rangle = ||x_n||^2-1$$
Now $$||x_n||^2=||\psi_1+\frac1n\sum_k^n\psi_{k+1}||^2=1+\sum_{k=1}^n \frac1{n^2}=1+\frac1{n}$$
from orthogonality of $\psi_n$ and as such $||x_n-\psi_1|| \to 0$ as $n \to \infty$.
