Question

HELP PLEASE

Answer 1

Determine whether the series is convergent or divergent.
\[\sum_{n=1}^{\infty}\arctan5n\]
If it is convergent, find its sum. (If the quantity diverges, enter DIVERGES.) __________

Answer 2

Arctan is bounded within \[\left[ -\frac{ \pi }{ 2 },\frac{ \pi }{ 2 } \right]\] right?

Answer 3

well

Answer 4

tan theta = opp/adj
theta = = arctan(opp/adj
^--- theta can actually revolve as many times as u want

Answer 5

so it diverges

Answer 6

Eh. Idunno how that works. D:

Answer 7

Oh wait. I think i get it

Answer 8

There is another problem that is\[\sum_{n=1}^{\infty} (\cos14)^k\]

Doesnt that have infinite revolutions too?

Answer 9

wait i wanna go over that again

Answer 10

so u know that if 5n is getting bigger that means opp/adj is getting larger

Answer 11

so this angle is increasing

Answer 12

since its strictly positive as n is from 1 to inf 5n is positive, so its bounded from 0 to pi/2

Answer 13

either way u are adding many number between 0 and pi/2 so it diverges

Answer 14

numbers

Answer 15

in this case cos(14) is just a ratio

Answer 16

Oh true, you can simply use the divergence test I suppose,\[\large\rm \lim_{n\to\infty}\arctan(5n)\quad=\quad\frac{\pi}{2}\]The `sequence` does not converge to zero, so the sum diverges.

Answer 17

since cos 14 < 1 it will converge
if cos 14 =1/r
then
sum k from 1 to inf (1/r)^k, this is a geom series

Answer 18

Oh I see now. Okay thanks!

Answer 19

yw ;)

Answer 20

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