Asymptotics for sum involving Euler numbers

Question

This first request may be easy, but the asymptotics for the next step has me scratching my head.

Through an informal inductive argument I have been able to show $$ (1) \quad \sum_{j=0}^{n-1}2^{2m(n-j)}E_{2j}\binom{2n}{2j} = 2\sum_{j=1}^{2^{m-1}}(-1)^j(2j-1)^{2n} , \quad m \ge 2.$$ The $E_{2j}$ are the Euler numbers. In trying to extend this to integers greater than two and I have acquired numerical evidence for $$ (2) \quad \sum_{j=0}^{n-1}r^{2m(n-j)}E_{2j}\binom{2n}{2j} \sim 2 (-1)^r\sum_{j=1}^{r^{m-1}}(-1)^j(rj-1)^{2n} , \quad m,r=2,3,...$$
I have searched for an appropriate identity but have only been able to postulate an asymptotic relationship. Note that because of the factor of $(-1)^r$ and the $r$ in the upper limit of the summation it is likely that (2) applies only to integer $r>1.$ A different proof than mine might help in my own investigations.

My question is: can (2) be proven, and if that is not possible, can (1) be proven without an induction; e.g., generating functions.

Answer 1

Let $$T_{k,n}(a,d) =a^k -(a+d)^k +(a+2d)^k +\cdots + (-1)^{n-1}\bigl(a+(n-1)d\bigr)^k.$$ Equation (7.4) of F. T. Howard, Sums of powers of integers via generating functions, Fibonacci Quarterly 34 (1996), no. 3, 244–256, is $$T_{k,n}(a,d) = \frac{d^k}{2}\left[(-1)^{n-1}E_k\left(\frac ad +n\right) +E_k\left(\frac ad\right) \right],\tag{$*$}$$ where $E_k(z)$ is the Euler polynomial defined by $$\frac{2e^{xz}}{e^x+1}=\sum_{k=0}^\infty E_k(z) \frac{x^k}{k!}.$$ In terms of the Euler numbers, we have $$E_k(z) = \sum_{i=0}^k \binom ki \frac{E_i}{2^i}\left( z-\frac12\right)^{k-i}$$ and the Euler numbers are given by $E_n = 2^n E_n (\tfrac12)$. (Note that $E_n=0$ for $n$ odd.) If we take $a=1, d=2$ in $(*)$ and simplify we get $$2\bigl(1^k -3^k +\cdots +(-1)^{n-1} (2n-1)^k \bigr)= (-1)^{n-1}\sum_{i=0}^{k-1}\binom ki E_i (2n)^{k-i}+\bigl((-1)^{n-1} +1\bigr) E_k,$$ which generalizes (1).

Taking $a=r-1, d=r$ in $(*)$ will give a formula for the right side of $(2)$ but it doesn't seem to be close to the left side of $(2)$.