Question

help please.. how do i find my testing points ?

please check my work

Answer 1

@zepdrix

Answer 2

zeppt come back D: lol

Answer 3

Grr remind me, what is radius of convergence?
Is it just half the length of the interval of convergence?

Answer 4

And what does that say in the bottom right corner of the page?
Find Miap+

Whut??

Answer 5

Ahh I looked it up :d yes, radius is just half.

I still don't know what "Find Miap+" means though lolol

Answer 6

oh to find the midpoint haha

Answer 7

Oh so that chicken scratch says Midp+ short for midpoint? XD
hahahaha

Answer 8

loool yah messy writing

Answer 9

I remember in the last problem we did the interval a bit different.
We excluded the end points and used square brackets.

So \(\large\rm I=(-2,3)\) would be the same as \(\large\rm I=[-1,2]\) as far as integer values go.
Are we supposed to use the rounded brackets?
Is that how they look in your other problems and stuff?

Answer 10

hmmm i think bc ive seen some that are for ex. (2 ,2]

Answer 11

Ok I'm confused then :[ ugh

Answer 12

I dunno, it seems like you did everything right to me lady >.<
I dunno you think you need my help!

Answer 13

lol yea :o me too

Answer 14

i do need ur help xD

Answer 15

its's okay we can skip this one lool

Answer 16

you have to test the end points
x=-2,3 substitute these into the original series
\[\sum_{n=1}^{\infty} \frac{(2x-1)^n}{5^n \sqrt{n}} \]

x=-2
\[\sum_{n=1}^{\infty} \frac{(2 \color{red}{(-2)}-1)^n}{5^n \sqrt{n}} \]
\[\sum_{n=1}^{\infty} \frac{( -5)^n}{5^n \sqrt{n}} \] \[\sum_{n=1}^{\infty} \frac{(-1)^n( 5)^n}{5^n \sqrt{n}} \]
\[\sum_{n=1}^{\infty} \frac{(-1)^n }{ \sqrt{n}} \]
Alternation series test:
\[\sum_{n=1}^{\infty} \frac{(-1)^n }{ \sqrt{n}} = \lim_{n \rightarrow \infty } \frac{1}{\sqrt{n}} =0\]
Thus, series converges.

so it should be [-2,??
now check x=3

Answer 17

\[\sum_{n=1}^{\infty} \frac{(2x-1)^n}{5^n \sqrt{n}} \]
\[\sum_{n=1}^{\infty} \frac{(2 \color{red}{(3)}-1)^n}{5^n \sqrt{n}} \]\[\sum_{n=1}^{\infty} \frac{(6-1)^n}{5^n \sqrt{n}} \]\[\sum_{n=1}^{\infty} \frac{(5)^n}{5^n \sqrt{n}} \]\[\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \]
P-series
P <1
series diverge
Therefore, IOC: [-2,3)
[ converges , diverge )
(diverge , Converges ] , (diverge , diverge )
[converges , converges]