Determine whether the sequence converge or diverges, and if it converges find the limit. $a_n=(4+\sin{n\pi\over2})$
I know that
$$a_n=5,4,3,4,5,4,3,4,\ldots$$
So $a_{2n} = 4,4,4,\ldots \to 4$
$$a_k=3,3,3,\ldots \to 3$$
Or $a_t=5,5,5,\ldots \to 5$
So $a_n$ diverges, but how can I find $k$ or $t$ by $n$ ?
$b_n =n\sin {1\over n}$
My attempt:
$$t= {1\over n}, t\to 0 \text{ when } n\to \infty.$$
So
$$\lim_{n \to \infty} n \sin {1\over n} = \lim_{t \to 0} {\sin t\over t}=1$$
$$c_n= {{100n}\over{n^{{3}\over{2}}+4}}$$
Let $t=\sqrt{n}$, $t\to \infty$ when $n\to \infty$.
$\lim_{n \to \infty} {{100n}\over{n^{{3}\over{2}}+4}}= \lim_{t \to \infty} {{100t^2}\over{t^{3}+4}}=0$
Is that true ? Thanks.
For $a$, try:
$$t= 4n+1, k= 4n+3$$
The others look fine.