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anonymous
 4 years ago
i have the sequence a_n=( sin 1!/1*2)+( sin 2!/2*3)+( sin 3!/3*4) + ... + ( sin n!/n*(n+1)) i must use cauchy (a_(n+p)  a_n) to verify if it converges or not
anonymous
 4 years ago
i have the sequence a_n=( sin 1!/1*2)+( sin 2!/2*3)+( sin 3!/3*4) + ... + ( sin n!/n*(n+1)) i must use cauchy (a_(n+p)  a_n) to verify if it converges or not

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ a_n=\frac{ sin (1!)}{1*2}+\frac{ sin (2!)}{2*3}+....+\frac{ sin( n!)}{n*(n+1)} \] This?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@henpen help me please :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can ignore the sine part (i think) since it is bounded above by 1and below by 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[a_{n+p}a_{n}=\frac{\sin((n+1)!)}{(n+1)(n+2)}+\frac{\sin((n+2)!)}{(n+2)(n+3)}+\cdots+\frac{\sin((n+p)!)}{(n+p)(n+p+1)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[a_{n+p}a_{n}\leq \frac{1}{(n+1)(n+2)}+\frac{1}{(n+2)(n+3)}+\cdots+\frac{1}{(n+p)(n+p+1)}\] which for n big enough is less than any epsilon

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yup, thak you very much
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