Question

I believe this is correct...


\[\sum_{n=1}^{\infty}\frac{x^{n}}{5^{n}n^{5}}\]
\[\lim_{n\rightarrow \infty} \frac{x^{n+1}}{5^{n+1}(n+1)^{5}} \frac{5^{n}n^{5}}{x^{n}}\]
\[\lim_{n\rightarrow \infty} \frac{x^{n+1}5^{n}}{x^{n}5^{n+1}} \frac{n^{5}}{(n+1)^{5}}\]
\[\lim_{n\rightarrow \infty} \frac{x}{5} \frac{n^{5}}{(n+1)^{5}}\]
\[\frac{1}{5}\lim_{n\rightarrow \infty} \left|x\right| \frac{n^{5}}{(n+1)^{5}}\]
\[\frac{1}{5}\lim_{n\rightarrow \infty} \left|x\right| \frac{1}{(1+1/n)^{5}}\]
\[\frac{1}{5}\lim_{n\rightarrow \infty} \left|x\right| \frac{1}{(1+1/n)^{5}}<1\]

Answer 1

R=5

Answer 2

\[\left|x\right|<5\]

Answer 3

I= [-5,5]

Answer 4

Radius of convergence and interval of convergence

Answer 5

what do you think @agentx5

Answer 6

Did you try the ratio test? This looks a lot like a combination series variant btw, a big hint to use the ratio test typically. :-D

Answer 7

yes I did the ratio test

Answer 8

\[\frac{a_{n+1}}{a_{n}}\]

Answer 9

Ah wait I see you posted it later, ack time delay.

Well since we know by the ratio test, the series converges when abs(x)<5, we also know the radius of convergence.

What you need now is to find the interval of convergence. We’ll get most (if not all) of the interval in question by solving the inequality from above. Do you remember how to write that as a three-part inequality? (two inequality signs)

Answer 10

I believed it was [-5,5]

Answer 11

about the three part in equality ...can you show me...i'm not sure

Answer 12

Right that's your radius :D

But you needed the interval too right?

Answer 13

Sure! one sec...

Answer 14

\[-5<x<5\]

Answer 15

For the less than, greater than would be...

Answer 16

\[x<-5 \cup x>5\]

Answer 17

so it would make more sense to write -5<x<5 instead of [-5,5]

Answer 18

It does, and I would get in the practice of it because you might end up with something like this:

\[ \left| x+3 \right| < 4\]

Answer 19

Which is: -7<x<1

Answer 20

ok what does \[x<-5 \cup x>5\] mean again

Answer 21

After you do this and find this inequality, or union of inequalities, you want to determine if the power series will converge or diverge at the endpoints of this interval.

Answer 22

@MathSofiya Remember the number line stuff you did way back when you were starting Algebra?

Answer 23

Disjunction and conjunction?

Answer 24

yes

Answer 25

set of all x's less than -5 "union" set of all x's greater than +5

Answer 26

Union is a fancy way of saying, "yeah both of these at the same time are true"

Answer 27

i see

Answer 28

That. lol

Answer 29

Versus the in-between with:

Answer 30

to determine if the power series will converge or diverge at the endpoints of this interval. do I use the initial equation? or this
\[\frac{1}{5}\lim_{n\rightarrow \infty} \left|x\right| \frac{1}{(1+1/n)^{5}}\]

Answer 31

You'll simply plug them at these points into the original power series and see if the series converges or diverges using any test necessary, to determine convergence. Make sense?

Answer 32

yes sir

Answer 33

Thanks!

Answer 34

PS: Bump this so the others notice it and we get some additional reviewers :-)

Answer 35

Did the answer check out ok?

Answer 36

agent, can you help cal and on with a geometry problem once yo uare done?