Computing trace of matrix Let A be a n×n matrix such that $$[a_{ij}]_{n×n}=\frac{((-1)^i)(2i^2+1)}{4j^4+1}$$ then what is
$$1+ \lim_{n\to {\infty}}\left(tr(A^n)^{1/n}\right)$$
I cannot figure out how to calculate trace of $$A^n$$
 A: If $\{\lambda_k\}$ are the eigenvalues of $A$, then $A^m$ has eigenvalues $
\{\lambda_k^m\}$ and so:
$$\text{trace} A^m = \sum_{i=1}^n\lambda_k^m.$$
If these eigenvalues are non-negative reals, then you can easily show that:
$$\lim\limits_{m\to\infty}\sqrt[m]{\lambda_1^m+\dots \lambda_n^m} = \max\limits_k \lambda_k.$$
what remains to be done is to find the eigenvalues of your matrix which I do not see an immediate formula for.
A: Answer complementing the answer of @dezdichado.
In fact this matrix is the product of the (column) vector $U$ with coordinates:
$$(-1)^i(2i^2+1), \ \ \ \ i=1 \cdots n$$
with the line vector $V^T=$ with coordinates
$$\frac{1}{4j^4+1}, \ \ \ \ j=1 \cdots n$$
Therefore $A=UV^T$ is a rank 1 matrix. As such, it has eigenvalue $0$ with multiplicity $n-1$, and a nonzero eigenvalue that is plainly the trace, therefore the real number:
$$\lambda=\sum_{i=1}^n a_{ii}=\sum_{i=1}^n \frac{(-1)^i(2i^2+1)}{4i^4+1}$$
which has a closed form expression due to its alternate signs yielding a telescoping behavior (observation made by Dezdichado):
$$\lambda=\dfrac{-2n^2-2n-1+(-1)^n}{(2n+1)^2+1}$$
Remark: the result obtained is the square of the so-called spectral norm of $A$.
