这是一份ucl伦敦大学学院 math0042作业代写的成功案
$1 / 4$ because $1 /(n+1)$ results from integration, and for the series without this factor in the coefficients we have in (6), Sec. 14.2,
$$
\frac{(2 n) ! /(n !)^{2}}{(2 n+2) ! /((n+1) !)^{2}}=\frac{(n+1)^{2}}{(2 n+1)(2 n+2)} \rightarrow \frac{1}{4} .
$$
$\infty$ because $3 n(3 n-1)$ results from differentiation, and for the coefficients without these factors we have in (6), Sec. 14.2,
$$
\frac{1 / n^{n}}{1 /(n+1)^{n+1}}=\left(\frac{n+1}{n}\right)^{n}(n+1) \quad \rightarrow \quad \text { as } n \rightarrow \infty
$$
MATH0042 COURSE NOTES ：
First of all, since $\sin (w+2 \pi)=\sin w$ and $\sin (\pi-w)=\sin w$, we obtain all values of $\sin w$ by letting $w$ vary in a suitable vertical strip of width $\pi$, for example, in the strip $-\pi / 2 \leqq u \leqq \pi / 2$. Now since
$$
\sin \left(\frac{\pi}{2}-i y\right)=\sin \left(\frac{\pi}{2}+i y\right)=\cosh y
$$
$$
\sin \left(-\frac{\pi}{2}-i y\right)=\sin \left(-\frac{\pi}{2}+i y\right)=-\cosh y
$$