Limit $\lim_{k\to\infty}\left(\sum_{r=1}^{k-1}\zeta\left(2r\right)\frac{\left(-1\right)^{r+k}}{\left(2k-2r-1\right)!}\right)$

Question

I am interested in finding this limit and the answer seems to be: $$\lim_{k\to\infty}\left(\sum_{r=1}^{k-1}\zeta\left(2r\right)\frac{\left(-1\right)^{r+k}}{\left(2k-2r-1\right)!}\right)=-\sin1$$ I kind of have a method for this but that is based on how I came across this series.

I would love to see other solutions that are not based on how it was derived.

EDIT:
Also as a side note it seems that, $$\lim_{k\to\infty}\left(\sum_{r=1}^{k-1}\frac{\left(-1\right)^{k+r}}{\left(2k-2r-1\right)!}\right)=-\sin1$$ The series without $\zeta$ equals the same.

EDIT 2:
I realized we could first simplify a bit by $r\to k-1+1-r=k-r$ $$\lim_{k\to\infty}\left(\sum_{r=1}^{k-1}\zeta\left(2k-2r\right)\frac{\left(-1\right)^{r}}{\left(2r-1\right)!}\right)=-\sin1$$ Also notice that, $$\lim_{k\to\infty}\left(\sum_{r=1}^{k-1}\frac{\left(-1\right)^{r}}{\left(2r-1\right)!}\right)=-\sin1$$ So I thought lets try it on other series,

Numerically, $$\lim_{k\to \infty}\left(\sum_{r=0}^{k}\frac{\zeta\left(2k-2r\right)}{r!}\right)=e$$

Also, notice that. $$\lim_{k\to \infty}\left(\sum_{r=0}^{k}\frac{1}{r!}\right)=e$$ This is kind of interesting, shouldn't the $\zeta$ term affect the summation?

Answer 1

ORIGINAL ANSWER

To put it succinctly. \begin{align*} \lim_{N\rightarrow \infty} \sum_{n=1}^{N-1}{\zeta \left( 2\left( N-n \right) \right) \frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}} =&\lim_{N\rightarrow \infty} \sum_{n=1}^{N-1}{\sum_{k=1}^{\infty}{\frac{k^{2n}}{k^{2N}}}\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}} \\ =&\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=1}^{N-1}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1} \\ =&\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1} \\ =&-\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{\sin \left( k \right)}{k^{2N-1}}}=-\sin \left( 1 \right) \end{align*} If you found this a bit too handwavy, I could provide the solution in greater detail.

EDIT

I shall provide a more rigorous proof here. To start with, we rewrite the sum in terms of the definition of riemann zeta function. So we have $$ \sum_{n=1}^{N-1}{\zeta \left( 2\left( N-n \right) \right) \frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}=\sum_{n=1}^{N-1}{\sum_{k=1}^{\infty}{\frac{k^{2n}}{k^{2N}}}\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}} $$ We can interchange the sums if each terms in the finite sum converges, which is obvious in this case. Therefore $$ \sum_{n=1}^{N-1}{\sum_{k=1}^{\infty}{\frac{k^{2n}}{k^{2N}}}\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}=\sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=1}^{N-1}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1} $$ Now we have to proof that $$ \lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=N}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}=0 $$ Since we have $$ \sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}=\sin \left( k \right) $$ Which converges, by something Cauchy (Cannot remember the exact name) $\forall \varepsilon >0, \exists N\in \mathbb{N}\,\,\mathrm{s}.\mathrm{t}.$ \begin{align*} &\sum_{n=N}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}<\frac{\varepsilon}{\zeta \left( 2 \right)}<\frac{\varepsilon}{\zeta \left( 2N-1 \right)} \\ \Rightarrow &\sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=N}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}<\sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\frac{\varepsilon}{\zeta \left( 2N-1 \right)}=\varepsilon \end{align*} Which proves the claim. Hence \begin{align*} &\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=1}^{N-1}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1} \\ =&\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=1}^{N-1}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}+0 \\ =&\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=1}^{N-1}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}+\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=N}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1} \\ =&\lim_{N\rightarrow \infty} \left( \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=1}^{N-1}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1}+\sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=N}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1} \right) \\ =&\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{\left( 2n-1 \right) !}}k^{2n-1} \\ =&-\lim_{N\rightarrow \infty} \sum_{k=1}^{\infty}{\frac{\sin \left( k \right)}{k^{2N-1}}} \end{align*} Now, notice that $$ \sum_{k=1}^{\infty}{\left| \frac{\sin \left( k \right)}{k^{2N-1}} \right|}\leqslant \sum_{k=1}^{\infty}{\frac{1}{k^{2N-1}}}=\zeta \left( 2N-1 \right) $$ Meaning the sum converge absolutely, implying the sum converges. So we only need to prove $$ \lim_{N\rightarrow \infty} \left| \sum_{k=2}^{\infty}{\frac{\sin \left( k \right)}{k^{2N-1}}} \right|=\lim_{N\rightarrow \infty} \left| \sin \left( 1 \right) -\sum_{k=1}^{\infty}{\frac{\sin \left( k \right)}{k^{2N-1}}} \right|=0 $$ The gist to prove this is along these lines \begin{align*} &\forall \varepsilon >0, \exists N\in \mathbb{N} \,\, \mathrm{s}.\mathrm{t}. \\ &\left| \sum_{k=2}^{\infty}{\frac{\sin \left( k \right)}{k^{2N-1}}} \right|\leqslant \left| \sum_{k=2}^N{\frac{\sin \left( k \right)}{k^{2N-1}}} \right|+\left| \sum_{k=N+1}^{\infty}{\frac{\sin \left( k \right)}{k^{2N-1}}} \right|\leqslant \sum_{k=2}^N{\frac{1}{k^N}}+\sum_{k=N+1}^{\infty}{\frac{1}{k^2}}<\sum_{k=2}^N{\frac{\varepsilon}{2^{k-1}}}+\frac{\varepsilon}{2^N}=\varepsilon \end{align*} By utilizing that cauchy stuff (name forgotten) and the basic definition of limits.
Then, we are done.

Answer 2

For the first one, I am stuck.

The second one is quite simple since

$$S_k(x)=\sum_{r=1}^{k-1}\frac{\left(-1\right)^{k+r}}{\left(2k-2r-1\right)!} x^r$$is $$S_k(x)=(-1)^k \left((-x)^{k-\frac{1}{2}} \sinh \left(\frac{1}{\sqrt{-x}}\right)-\frac{\, _1F_2\left(1;k,k+\frac{1}{2};-\frac{1}{4 x}\right)}{(2 k-1)!}\right)$$ $$S_k(1)=-\sin(1)-\frac{(-1)^k \, _1F_2\left(1;k,k+\frac{1}{2};-\frac{1}{4}\right)}{\Gamma (2 k)}$$

Notice that for $k>0$ $$\sqrt{\frac{\pi }{2}} J_{\frac{1}{2}}(1)\leq \, _1F_2\left(1;k,k+\frac{1}{2};-\frac{1}{4}\right) < 1$$ expalins the very fast convergence.

Neglecting the hypergeometric function $$S_k(x) \sim -x^{k-\frac{1}{2}} \sin \left(\frac{1}{\sqrt{x}}\right)$$