Questions about $Ax:=\sum_{n=1}^\infty \lambda_n(x,e_n)e_n$ I want to show the following:

Let $X$ be a seperable Hilbertspace, $(e_n)_{n\in\mathbb{N}}$ an orthonormal basis of $X$ and $(\lambda_n)_{n\in\mathbb{N}}\subset\mathbb{R}$ a bounded sequence.
$i)$ $\sum_{n=1}^\infty \lambda_n(x,e_n)e_n$ converges in $X\nobreakspace \forall x\in X$
$ii)$ $A:X\rightarrow X, Ax:=\sum_{n=1}^\infty \lambda_n(x,e_n)e_n$ is linear, continuous and self adjoint

What I've shown already is that A is linear and continuous, but I'm stuck on self adjoint. I know that self adjoint means $A=A$* with $A$* being the Hilbertspace-adjoint. But how do I show that without knowing how $A$* looks like? I have problems with $i)$ as well. I tried different techniques, but none of them seem to work. Can someone help me? Thanks in advance.
 A: Let $c=\sup\{|\lambda_n|:\ n\}$. You have, since $\{e_n\}$ is orthonormal, 
\begin{align}
\left\|\sum_{n=r}^m\lambda_n\langle x,e_n\rangle\,e_n\right\|^2
&=\sum_{n=r}^m|\lambda_n|^2\,|\langle x,e_n\rangle|^2
\leq c^2\sum_{n=r}^m|\langle x,e_n\rangle|^2.
\end{align}
Since $\sum_n|\langle x,e_n\rangle|^2=\|x\|^2$ (Parseval), the above shows that the sequence of partial sums of your series is Cauchy; so the series converges. 
For the adjoint: I wrote conjugates because the computation lest you calculate $A^*$ even when the $\lambda_n$ are not real. Only in the last step have I used that $\lambda_n$ is real. 
\begin{align}
\langle A^*x,y\rangle&=\langle x,Ay\rangle
=\left\langle \sum_n\langle x,e_n\rangle e_n,\sum_m\lambda_m\langle y_m,e_m\rangle\,e_m\right\rangle\\ \ \\
&=\sum_{m,n}\overline{\lambda_m}\,\langle x,e_n\rangle\,\overline{\langle y,e_m\rangle}\,\langle e_n,e_m\rangle=\sum_n\overline{\lambda_n}\,\langle x,e_n\rangle\,\overline{\langle y,e_n\rangle}\\ \ \\
&=\sum_{m,n}\overline{\lambda_n}\,\langle x,e_n\rangle\,\overline{\langle y,e_m\rangle}\,\langle e_n,e_m\rangle\\ \ \\
&=\left\langle \sum_n\overline{\lambda_n}\langle x,e_n\rangle e_n,\sum_m\langle y_m,e_m\rangle\,e_m\right\rangle\\ \ \\
&=\langle Ax,y\rangle.
\end{align}
