I'm struggling with following exercise (Ex.9) from the book in the corresponding chapter.
Using Lemma 1 show the radius of convergence of $\sum B_nt^n/n!$ is $2\pi$. As a consequence show that for any $C,k>0$ there are infinitely many $n$ such that $|B_n|>Cn^k$.
Lemma 1 simply says that $$\frac{t}{e^t-1}=\sum_{n=0}^\infty B_nt^n/n!$$
I managed to show $$\limsup_{n\to\infty}\sqrt[n]{\frac{|B_n|}{n!}}=2\pi$$ using identity $2(2m)!\zeta(2m)=(-1)^{m+1}(2\pi)^{2m}B_{2m},$ which does not solves the problem (I mean I'm using stronger fact).
So how can I compute the $\limsup$ using only the Lemma?
I have a problem with the second part as well. From the $\limsup$ for every $\varepsilon>0$ there is $N$ such that $|B_n|/n!<2\pi+\varepsilon$ for every $n>N$. Let us assume $n$ even and fix $\varepsilon.$ Then we have bound $$|B_n|<n!(2\pi+\varepsilon).$$ We know $\forall C,k>0$ there is $M$ such that $Cn^k<n!$ for every $n>M$, so an obvious necessary condition is $n>\max\{N,M\}$. But what else?