Question

Help Please

Answer 1

Could you explain these two problems to me?

Answer 2

Both, or just the second?

Answer 3

\(\displaystyle \lim_{n\to \infty}\arctan(5n)=\pi/2\ne0 \).

As again, your sequence does not converge to zero.
(This is why you were correct on the first problem.)

Answer 4

Would you agree if I say that for any non-zero integer \(x\), \(\displaystyle |\cos x|<1\) ?

Answer 5

A maximum of a cosine (other than x=0) is always located at
\(\displaystyle x=\frac{a}{b}\pi\) where \(a\) and \(b\) are integers.

In other words (rational number) * pi.

(This is never true for an integer. In other words,
\(\displaystyle \frac{a}{b}\pi\ne \rm integer\) if \(a\) and \(b\) are integers.)

Answer 6

So, a cosine of an integer is just by definition less, because
cos(integer) is never the maximum of a cosine function
(and the maximum of a cosine is 1).

Thus, \(|\cos(14)|<1\).

Answer 7

And that now is a geometric series, or for the very least it can be compared to such (by comparison test).

Answer 8

if I lost you, let me know.

Answer 9

\(\displaystyle \sum_{k=1}^{\infty}(\cos 14)^k\)

You know that
\(\displaystyle \sum_{k=1}^{\infty}r^k\)
converges whenever \(\displaystyle |r|<1\).

Answer 10

oh forgot to add.

Also,
\(\displaystyle \sum_{k=1}^{\infty}r^k=\frac{r}{1-r}\)

Answer 11

If you want any further explanations or derivations, ask.
(If I am offline, I don't guarantee a reply)