Learning about the harmonic series divergence and p-harmonics series convergence, I tried to manipulate them globally as follows. $$\lim_{n \to \infty}A \cdot I = \lim_{n \to \infty} \begin{pmatrix} \frac{1}{1} & \frac{1}{2} & ... & \frac{1}{n} \\ \frac{1}{1^2} & \frac{1}{2^2} & ... & \frac{1}{n^2} \\ ... & ... & ... & ... \\ \frac{1}{1^n} & \frac{1}{2^n} & ...& \frac{1}{n^n} \\ \end{pmatrix}\cdot \begin{pmatrix} 1 \\ 1 \\ ... \\ 1 \end{pmatrix} = \begin{pmatrix}\sum_1^\infty\frac{1}{n} \\ \sum_1^\infty\frac{1}{n^2} \\ ... \\ \sum_1^\infty\frac{1}{n^n} \\ \end{pmatrix}= v $$ Where $A$ is the matrix of the harmonic and p-harmonic series, $I$ is the identity matrix and $v$ would be a vector of an infinite dimensional space.
One of the interesting things of this composition is that the trace of $A$ is curiously (please correct me if this is wrong) the Sophomore's Dream identity. And it is equivalent to the sum of the eigenvalues of the matrix.
$$tr(A)= \sum_1^{\infty} \frac{1}{n^n} = \int_0^1 x^{-x} dx = \sum_i \lambda_i$$
I would like to ask the following questions:
Are the matrix and vector operations correctly defined as I wrote them?
Topologically, how would look $v$ in that multidimensional space? initially all the dimensions converge to a number, because they depend of the p-harmonic series, but the dimension that depends on the harmonic-series does not converge.
The trace is also convergent, but $v$ is not totally convergent to a point (meaning it is not a point in that multidimensional space due to the dimension that depends on the harmonic series). Is that compatible?
In the other hand, in the case of the transpose matrix, $A^T\cdot I =v'$, if I am not wrong $v'$ would be totally convergent to a point in all its dimensions, so it would really be a point in the multidimensional space, is that intuition right? Thank you!