d) limnoo 1 •k=1 an e) limn 1 48x = =공 6 4
. Let $n \in \mathrm{N}^{*}$ and the equation:
$$
C_{2 n+1}^{1} x^{n}-C_{2 n+1}^{3} x^{n-1}+C_{2 n+1}^{5} x^{n-2}-\ldots+(-1)^{k} C_{2 n+1}^{2 k+1} x^{n-k}+\ldots+(-1)^{n} C_{2 n+1}^{2 n+1}=0
$$
Show that:
a) the equation admits as roots the numbers::
$\mathrm{x}_{\mathrm{k}}=\operatorname{ctg}^{2} \frac{k \pi}{2 n+1}$, for every $k \in\{1.2, \ldots . n\}$
b) $\frac{\pi^{2} n(2 n-1)}{3(2 n+1)^{2}}<\sum_{k=1}^{n} \frac{1}{k^{2}}<\frac{2 \pi^{2} n(n+1)}{3(2 n+1)^{2}}$
c) $\frac{\pi^{4} n(2 n-1)\left(4 n^{2}+10 n-9\right)}{45(2 n+1)^{4}}<\sum_{k=1}^{n} \frac{1}{k^{4}}<\frac{8 \pi^{4} n(n+1)\left(n^{2}+n+3\right)}{45(2 n+1)^{4}}$
d) $\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}$
e) $\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{k^{4}}=\frac{\pi^{4}}{4}$
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I need complete solution with explanation please.
the all fifth subpoints are for the same problem and they are conected between them, but if it is not possible to solve them all please solve what you can, thank you in advance.
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