Determine-if-the-series-n-1-a-n-defined-by-the-formula-converges-or-diverges-a-1-4-a-n-1-10-sin-n-n-a-n-

Question Number 140833 by liberty last updated on 13/May/21

Determine if the series \underset{n = 1}{\sum^{\infty}} a_{n}

defined by the formula converges or

diverges . a_{1} = 4, a_{n + 1} = \frac{10 + \sin n}{n} . a_{n}

Commented by liberty last updated on 13/May/21

can anyone help me

Answered by TheSupreme last updated on 13/May/21

\frac{a_{n + 1}}{a_{n}} = \frac{10 + s i n (n)}{n}

l i m \frac{a_{n + 1}}{a_{n}} = 0 \to s e r i e s c o n v e r g e s

Commented by liberty last updated on 13/May/21

why \lim \frac{a_{n + 1}}{a_{n}} = 0 ? as n \to ?

Answered by mathmax by abdo last updated on 13/May/21

a_{n + 1} = \frac{10 + \sin (n)}{n} a_{n} \Rightarrow \frac{a_{n + 1}}{a_{n}} = \frac{10 + \sin (n)}{n} \Rightarrow

\prod_{k = 1}^{n - 1} \frac{a_{k + 1}}{a_{k}} = \prod_{k = 1}^{n - 1} (\frac{10 + \sin (k)}{k}) \Rightarrow \frac{a_{2}}{a_{1}} . \frac{a_{3}}{a_{2}} \dots . . \frac{a_{n}}{a_{n - 1}} = \frac{1}{(n - 1)!} \prod_{k = 1}^{n - 1} (10 + \sin (k)) \Rightarrow

a_{n} = \frac{4}{(n - 1)!} \prod_{k = 1}^{n - 1} (10 + \sin (k)) \Rightarrow∣ a_{n} ∣⩽ \frac{4 . 11^{n}}{(n - 1)!} = v_{n}

\frac{v_{n + 1}}{v_{n}} = \frac{4 . 11^{n + 1}}{n!} \times \frac{(n - 1)!}{4 . 11^{n}} = \frac{11}{n} \to 0 (n \to \infty) \Rightarrow Σ v_{n} converges \Rightarrow

Σ a_{n} cv .

Answered by physicstutes last updated on 13/May/21

a_{n + 1} = \frac{10 + \sin n}{n} a_{n}

\Rightarrow \frac{a_{n + 1}}{a_{n}} = \frac{10 + \sin n}{n} = \frac{10}{n} + \frac{\sin n}{n}

\lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = \lim_{n \to \infty} (\frac{10}{n} + \frac{\sin n}{n}) = 0 + 0 = 0 < 1

\Rightarrow \underset{n = 1}{\sum^{\infty}} a_{n} converges by the ratio test .