Question

does 1/n*[sin(n)+cos(n)] converge to a specific value (if not, why not)?

Answer 1

so is it 1/(n* (sin(n) + cos(n)) or is it (1/n) * (sin (n) + cos(n))

Answer 2

(1/n) * (sin (n) + cos(n)), because of the operator precedence

Answer 3

Is this limit as n goes to infinity?

Answer 4

So now it is clear what you mean. What is your own opinion about it? What do you know about 1/n and sin n + cos n as n becomes large?

Answer 5

Assuming n goes to infinity ;)

Answer 6

Hint: write as \[\frac{ \sin n + \cos n }{ n }\]and then look at the absolute value: \[\frac{ |\sin n +\cos n| }{ n }\]Could you think of a number p so that |sin n + cos n| < p for every n?

Answer 7

@random_guy : it's your turn now ;)

Answer 8

@ZeHanz : thanks for the hint.

So it should contevrge to zero, not true? sin and cos return values betwen 0, 1, and because they are in pahse the maximal sum of the two is \[\sqrt{2}\]
As n approaches infinity the fraction gets smaller and smalle, but it's never 0 (it approches to it from both sides though)

Answer 9

You can take an easier upper bound for |sin n + cos n|, not worrying about the phase: \[0 \le \frac{ | \sin n+\cos n| }{ n }\le \frac{ 2 }{ n }\]
Now because 2/n has limit 0 as n goes to infinity, by the "squeeze theorem" (or how it is called) your expression also has limit 0.