I want to show that
$$\bigg \langle x-\sum_{i=1}^n \langle x,e_{i} \rangle e_{i} , e_{k} \bigg\rangle = 0$$
where $(e_{n}, n = 1, \dots, \infty)$ is an orthonormal basis of an inner product space $H$ and $x$ belongs to the space $H$. It starts simply
$$\bigg \langle x - \sum_{i=1}^{n} \langle x, e_{i} \rangle e_{i}, e_{k} \bigg \rangle = \langle x, e_{k} \rangle - \bigg\langle \sum_{i=1}^{n} \langle x, e_{i} \rangle e_{i}, e_{k} \bigg\rangle$$
My question occurs here. In the second part of the equation above, $$\bigg \langle \sum_{i=1}^{n} \langle x, e_{i} \rangle e_{i}, e_{k} \bigg \rangle$$
How do I treat this? Can I get the $\sum_{i=1}^n \langle x, e_{i} \rangle$ out of the inner product so then I'll have $$\sum_{i=1}^{n} \langle x, e_{i} \rangle \cdot \langle e_{i}, e_{k} \rangle$$
I assume that $\sum_{i=1}^n \langle x, e_{i} \rangle$ is a real number? I can't understand.
Thanks in advance for your help and sorry for any mistakes I might have done in typing this.
The inner product is linear in its first argument so:
\begin{align} \left\langle \sum_{i=1}^{n} \langle x, e_{i} \rangle e_{i}, e_{k} \right\rangle &= \bigg\langle \langle x, e_{1} \rangle e_{1} + \langle x, e_{2} \rangle e_{2} + \cdots + \langle x, e_{n} \rangle e_{n}, e_{k} \bigg\rangle\\ &= \langle x, e_{1} \rangle \langle e_{1}, e_k\rangle + \langle x, e_{2} \rangle \langle e_{2}, e_k\rangle + \cdots + \langle x, e_{n} \rangle \langle e_{n}, e_{k} \rangle\\ &= \langle x, e_{1} \rangle \delta_{1k} + \langle x, e_{2} \rangle \delta_{2k} + \cdots + \langle x, e_{n} \rangle \delta_{nk}\\ &= \langle x, e_k\rangle \delta_{kk}\\ &= \langle x, e_k\rangle \end{align}
where $\delta_{ij}$ is the Kronecker delta. Therefore
$$\left \langle x - \sum_{i=1}^{n} \langle x, e_{i} \rangle e_{i}, e_{k} \right \rangle = \langle x, e_{k} \rangle - \left\langle \sum_{i=1}^{n} \langle x, e_{i} \rangle e_{i}, e_{k} \right\rangle = \langle x, e_{k} \rangle - \langle x, e_{k} \rangle = 0$$