Use the Squeeze Theorem to prove that each of the following sequences converges :?

Question

a) {sin(n/n)}

b) {(n^3)/(2^n)}

c) {(3^n)/n!}

Answer 1

a) the converges to sin(1)
b) 1/2^n<(n^3)/(2^n) < n^3 / n^4
last < because 2^n>n^4 for n>5
left and right bounds -> zero , mid is squeezed between

c) 1/n!< (3^n)/n! < 3^n/4^n 0.75^n
agin n ! > 4^n for n >5 or 6 dont know , n big enough,



now was this that difficult that you had to ask the whole mathematical society ?

Answer 2

a) I think you meant (sin n)/n? (That is, [sin(n)]/n.)

This is easy enough. What is the range of the sine function? It is -1 <= sin n <= 1, where <= means less than or equal. (Think of it as an underlined < sign.)

Now divide that inequality by n, which is always positive in this case. And what are the limits of {-1/n} and {1/n}?