In the linear space of sequences $x=( x_{1}, x_{2}, ... ), (x_{k}\in \mathbb{R})$ such that $\sum_{k=1}^{\infty }x_{k}^{2}<\infty$. Let $(x,y)=\sum_{k=1}^{\infty }\lambda_{k}x_{k}y_{k}$, where $\lambda_{k}\in \mathbb{R}, 0<\lambda_{k}<1$. Will the resulting Euclidean space be Hilbertian?
Answer:
No. Let sequence $\sum_{k=1}^{\infty }\lambda_{k}$ is converge and for choosing $\varepsilon>0$ number $N$ such that $\sum_{k=N+1}^{\infty }\lambda_{k}<\varepsilon$. Let's choose sequence $x_{n}=(1,..,1,0,0 ... )$ with $n$ ones. Then if $m,n>N$ we have that $ \left \| x_{m} - x_{n} \right \|<\varepsilon $ but $x_{n}$ doesn't converge.
Questions: Why $x_{n}$ doesn't converge and why $ \left \| x_{m} - x_{n} \right \|<\varepsilon$? And why this is enough for proving?
You already asked this question and I already answered it here: Does the sequence $x_{n}=(1, ... , 1, 0,0,...)$ (with $n$ ones) converge?
Well, technically, it should be $\sqrt{\epsilon}$, but that doesn't change the proof. You can see why it's $\sqrt{\epsilon}$ if you write out
$$\|x_m-x_n\| = \|(0,\dots,0,1,\dots,1,0,\dots)\| = \sqrt{((0,\dots,0,1,\dots,1,0,\dots),(0,\dots,0,1,\dots,1,0,\dots))}$$
Because you have proven that there exists a non-convergent Cauchy sequence, therefore the statement "Every Cauchy sequence is convergent" (i.e. the definition of a Hilbert space) is false.