Question

HELP PLS.

Answer 1

Through Comparison,
\[\frac{ 1 }{ n^{0.78}} < \frac{ 1 }{ n }\]
So does that mean it diverges?

Answer 2

if i remember correctly the critical point for convergence for 1/a^n is when n=1

1/a^n when n=2 converges to 0 where 1/a divergers

Answer 3

yes

Answer 4

Wait. Doesnt \[\frac{ 1 }{ n^{0.78} } >\frac{ 1 }{ n }\]
for it to diverge?

Answer 5

Like I would have to find a similar fraction that is greater than the original for it to diverge. It doesnt work if its less than

Answer 6

It diverges because the exponent is not greater than 1.

Answer 7

\(\large\rm n^{0.78}<n^1\)

\[\large\rm \frac{1}{n^{0.78}}\cancel{<}\frac{1}{n^1}\]

Answer 8

\[\large \frac{ 1 }{ n^{0.78} } >\frac{ 1 }{ n }\]this is true, so it diverges.

Answer 9

Oh okay.

Answer 10

\(\Large \frac{ 1 }{ n^a }\) converges if a > 1.

Answer 11

Oh okay. I can picture it graphically as well. Thank you

Answer 12

So then Number 9 would be Convergent.

Answer 13

?

Answer 14

@zepdrix

Answer 15

Who what :U

Answer 16

Nevermind. i got it. lol

Answer 17

9 converges? Mmm ya that sounds right, probably.
The terms are shrinking fast enough since the exponent is more negative than -1.

Answer 18

Then how did i get this wrong?
Isnt that diverging? Since it is going really slow?

Answer 19

Python huh? :D Interesting

Answer 20

Oh yea. lul

Answer 21

\[\large\rm 1+\frac{1}{2\sqrt2}+\frac{1}{3\sqrt3}+...\]Can we write this in a more general way?
I guess the terms look like this, yes?\[\large\rm \frac{1}{n \sqrt n}\]Written another way,\[\large\rm \frac{1}{n\cdot n^{0.5}}\quad=\quad\frac{1}{n^{1.5}}\]

Answer 22

O.

Answer 23

Thanks!