Wyznacz granicę ciągu

[egi]

a) \(a_n\) = \(\sqrt[n]{ \frac{6^n-e^n-π^n}{3^n+e^n+π^n}}\)

b) \(b_n\) = \(\frac{2n+ \sqrt{1} }{n^2+ \sqrt{1} } + \frac{2n+ \sqrt{2} }{n^2+ \sqrt{2} } + ... + \frac{2n+ \sqrt{n} }{n^2+ \sqrt{n} }\)

c) \(c_n\) = \(\frac{\sqrt[3]{n^4}-1}{ \sqrt[3]{ 2n^7}-1} + \frac{\sqrt[3]{n^4}-2}{ \sqrt[3]{ 2n^7}-2} + ... + \frac{\sqrt[3]{n^4}-n}{ \sqrt[3]{ 2n^7}-n}\)

d) \(d_n\) = \(n* \sqrt{1- \frac{2}{n} } - n^2 * \sqrt{ \frac{4}{n^3} - \frac{2}{n^4} + \frac{1}{n^2} }\)



Potrafię wyznaczać granice ciągu takie jak np. \(x_n\) = \(\sqrt[n]{3^n + 5^n + 7^n}\) a w powyższych przykładach nawet nie wiem jak zacząć

[eresh]

a) \(a_n\) = \(\sqrt[n]{ \frac{6^n-e^n-π^n}{3^n+e^n+π^n}}\)

\(\Lim_{n\to\infty}\sqrt[n]{\frac{6^n(1-(\frac{e}{6})^n-(\frac{\pi}{6})^n)}{\pi^n((\frac{3}{\pi})^n+(\frac{e}{\pi})^n+1)}}=\frac{6}{\pi}\Lim_{n\to\infty}\sqrt[n]{\frac{1-(\frac{e}{6})^n-(\frac{\pi}{6})^n}{(\frac{3}{\pi})^n+(\frac{e}{\pi})^n+1)}}=\frac{6}{\pi}\cdot 1=\frac{6}{\pi}\)

[eresh]

d) \(d_n\) = \(n* \sqrt{1- \frac{2}{n} } - n^2 * \sqrt{ \frac{4}{n^3} - \frac{2}{n^4} + \frac{1}{n^2} }\)

\(\Lim_{n\to\infty}n \sqrt{1- \frac{2}{n} } - n^2 \sqrt{ \frac{4}{n^3} - \frac{2}{n^4} + \frac{1}{n^2} }=\Lim_{n\to\infty}(\sqrt{n^2-2n}-\sqrt{4n-2+n^2})=\Lim_{n\to\infty}\frac{n^2-2n-4n+2-n^2}{\sqrt{n^2-2n}+\sqrt{4n-2+n^2}}=\\=\Lim_{n\to\infty}\frac{-6n+2}{n(\sqrt{1-\frac{2}{n}}+\sqrt{\frac{4}{n}-\frac{2}{n^2}+1})}=\Lim_{n\to\infty}\frac{-6+\frac{2}{n}}{(\sqrt{1-\frac{2}{n}}+\sqrt{\frac{4}{n}-\frac{2}{n^2}+1})}=\frac{-6}{2}=-3\)

[radagast]

b) \(b_n\) = \(\frac{2n+ \sqrt{1} }{n^2+ \sqrt{1} } + \frac{2n+ \sqrt{2} }{n^2+ \sqrt{2} } + ... + \frac{2n+ \sqrt{n} }{n^2+ \sqrt{n} }\)

\(\frac{2n+ \sqrt{1} }{n^2+ \sqrt{n} } + \frac{2n+ \sqrt{2} }{n^2+ \sqrt{n} } + ... + \frac{2n+ \sqrt{n} }{n^2+ \sqrt{n} }\le b_n \le \frac{2n+ \sqrt{1} }{n^2+ \sqrt{1} } + \frac{2n+ \sqrt{2} }{n^2+ \sqrt{1} } + ... + \frac{2n+ \sqrt{n} }{n^2+ \sqrt{1} }\)
czyli
\(\frac{2n+ \sqrt{1}+2n+ \sqrt{2}+...+2n+ \sqrt{n} }{n^2+ \sqrt{n} }\le b_n \le \frac{2n+ \sqrt{1}+2n+ \sqrt{2}+...+2n+ \sqrt{n} }{n^2+ \sqrt{1} }\)
czyli
\(\frac{2n^2+ n}{n^2+ \sqrt{n} }\le\frac{2n^2+ \sqrt{1}+ \sqrt{2}+...+ \sqrt{n} }{n^2+ \sqrt{n} }\le b_n \le \frac{2n^2+ \sqrt{1}+ \sqrt{2}+...+ \sqrt{n} }{n^2+ \sqrt{1} } \le \frac{2n^2+ n \sqrt{n} }{n^2+ \sqrt{1} }\)
tymczasem
\(\Lim_{n\to \infty }\frac{2n^2+ n}{n^2+ \sqrt{n} }=2= \Lim_{n\to \infty }\frac{2n^2+ n}{n^2+ 1 }\)
no to \(\Lim_{n\to \infty }b_n=2\)

[radagast]

c) \(c_n\) = \(\frac{\sqrt[3]{n^4}-1}{ \sqrt[3]{ 2n^7}-1} + \frac{\sqrt[3]{n^4}-2}{ \sqrt[3]{ 2n^7}-2} + ... + \frac{\sqrt[3]{n^4}-n}{ \sqrt[3]{ 2n^7}-n}\)

\(\frac{\sqrt[3]{n^4}-1}{ \sqrt[3]{ 2n^7}-1} + \frac{\sqrt[3]{n^4}-2}{ \sqrt[3]{ 2n^7}-1} + ... + \frac{\sqrt[3]{n^4}-n}{ \sqrt[3]{ 2n^7}-1}\le
c_n \le
\frac{\sqrt[3]{n^4}-1}{ \sqrt[3]{ 2n^7}-n} + \frac{\sqrt[3]{n^4}-2}{ \sqrt[3]{ 2n^7}-n} + ... + \frac{\sqrt[3]{n^4}-n}{ \sqrt[3]{ 2n^7}-n}\)
czyli
\(\frac{n\sqrt[3]{n^4}-(1+2+...+n)}{ \sqrt[3]{ 2n^7}-1}\le
c_n \le
\frac{n\sqrt[3]{n^4}-(1+2+...+n)}{ \sqrt[3]{ 2n^7}-n}\)
czyli
\(\frac{n\sqrt[3]{n^4}- \frac{n(1+n)}{2} }{ \sqrt[3]{ 2n^7}-1}\le
c_n \le
\frac{n\sqrt[3]{n^4}- \frac{n(1+n)}{2} }{ \sqrt[3]{ 2n^7}-n}\)
czyli
\(\frac{2n^ \frac{7}{3} - n-n^2}{2 \sqrt[3]{ 2}n^ \frac{7}{3} -2}\le
c_n \le
\frac{2n^ \frac{7}{3} - n-n^2}{2 \sqrt[3]{ 2}n^ \frac{7}{3} -2n}\)
...
na moje oko wychodzi
\(\Lim_{n\to \infty } c_n= \frac{1}{ \sqrt[3]{2} }\)
(ale mogę się mylić -lepiej to sprawdź)