Question

how to determine if its converges or diverges

Answer 1

\[\frac{ \cos \left(\begin{matrix}\frac{ n \pi }{ 2 } \\ \end{matrix}\right) }{ \sqrt{n} }\]

Answer 2

\[-1\le cosn \le1\]

Answer 3

or is it \[\frac{ -1 }{ \sqrt{n} } \le \cos \frac{ \frac{ n \pi }{ 2} }{ \sqrt{n}}\le \frac{ 1 }{ \sqrt{n} }\]

Answer 4

\[\large\rm -1\le \cos\left(\frac{n \pi}{2}\right)\le1\]You start here, ya?
And then divide "each side" by sqrt(n).

Answer 5

\[\large\rm \frac{-1}{\sqrt n}\le \frac{\cos\left(\frac{n \pi}{2}\right)}{\sqrt n}\le\frac{1}{\sqrt n}\]

Answer 6

\[\large\rm \lim_{n\to\infty}\frac{1}{\sqrt n}=?\]

Answer 7

Nice work, zepdrix. Note: zepdrix is using the so-called Squeeze Theorem to determine whether or not the given quantity converges.

Answer 8

yea that what i was thinking but not sure if i should start with cosn first in the middle

Answer 9

\[\frac{ 2\tan^{-1}n }{ n^{3}+4 }\]

Answer 10

can you help me on this one also?

Answer 11

\[\large\rm \frac{ 2\color{orangered}{\arctan(n)}}{n^{3}+4}\]
Hmm I think this one will work out the same way,
just with a different end value.
If you remember what the graph of arctangent looks like, that helps.It has a horizontal asymptote at pi/2.