Question

convergence of series

Answer 1

I'm stuck on a certain part

Answer 2

\[\sum_{n=1}^{\infty} (-1)^{n} \frac{ 2^{n} n! }{ n^{n} }\]

Answer 3

I used the ratio test and simplified it down to \[\frac{ 2(n+1) }{ (n+1) }*\frac{ n^{n} }{ (n+1)^{n} }\]

Answer 4

I know the first part cancels two and the second part the bottom is e. Is there anything I can do with the top or is it just infinity?

Answer 5

\[2 \lim_{x \rightarrow \infty}(\frac{1}{x+1})^x =2 \lim_{x \rightarrow \infty}(\frac{1}{x})^x=2 \lim_{x \rightarrow \infty}x^{-x}\]
can you find that last limit?

Answer 6

actually you could leave it as 1/x^x
1/infty =0

Answer 7

ratio test

Answer 8

oh, you used it already, :D

Answer 9

how did you get 1/x+1? isn't it x/x+1?

Answer 10

@freckles

Answer 11

oops that was suppose to be an x

Answer 12

i forgot about him

Answer 13

is it the same limit then? just a typo?

Answer 14

\[\lim_{x \rightarrow \infty}(\frac{x}{x+1})^x=\lim_{x \rightarrow \infty}(\frac{x+1-1}{x+1})^x=\lim_{x \rightarrow \infty} (1-\frac{1}{x+1})^x\]
see if you can take it from here :)

Answer 15

would it be -e then?

Answer 16

you mean e^(-1) ?

Answer 17

and don't forget to multiply your result by that 2 we had

Answer 18

no I do not. How would I be able to tell what e it is?

Answer 19

if you don't know you can always put it in the form of l'hospital
you know write it as e^ln( )

Answer 20

since in ratio test we are taking limit of the absolute value don't worry about (-1)^n part.

(this is simplified from the first set up)

\(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}\frac{2^{n+1}(n+1)!}{(n+1)^{n+1}}\times \frac{(n)^{n}}{2^{n}(n)!}}\)

\(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}\frac{n^n2(n+1)}{(n+1)^{n+1}}}\)

\(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}\frac{2n^n}{(n+1)^{n}}}\)

\(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}\frac{2n^n}{(n+1)^{n}}}\)

= 2/e

Answer 21

that is less than 1 and thus convergetnt

Answer 22

btw, love the way I am able to connect to the site... sorry it took so much time

Answer 23

@recon14193 have you tried that l'hospital?

Answer 24

\[(\frac{x}{x+1})^x=e^{ x \ln(\frac{x}{x+1})} \\ \text{ so \let's figure out the exponent part as x goes \to infinity } \\ e^{\lim_{x \rightarrow \infty}\frac{\ln(\frac{x}{x+1})}{\frac{1}{x}}}\]

you have the exponent is 0/0
so you can apply l'hosptial there

Answer 25

oh ok so that's how it is e^-1

Answer 26

well yeah because eventually we can show that exponent goes to -1

Answer 27

alright can I ask 1 more

Answer 28

by the e^(-1) is the same as 1/e

Answer 29

so you would end up with either 2*e^(-1) or 2/e as solomon said

Answer 30

what is the question

Answer 31

I just need help on which method to use

Answer 32

\[\sum \frac{ 10^{n} }{ (n+1)*4^{2n+1} }\]

Answer 33

I did ratio test and got 0

Answer 34

so I am not sure where to go from here

Answer 35

one sec I will try ratio test and see if I get that too

Answer 36

by the way L<1 which means it is convergent
if that really is what you got

Answer 37

I was told that also but if L=0 it means that were not sure what it is

Answer 38

no it is L=1

Answer 39

but anyways I don't get L=0

Answer 40

http://tutorial.math.lamar.edu/Classes/CalcII/RatioTest.aspx
see here if you don't believe me

Answer 41

thank you I see my mistake

Answer 42

\[\frac{10^{n+1}}{(n+2)4^{2(n+1)+1}} \cdot \frac{(n+1)4^{2n+1}}{10^n} \\ =\frac{10^n 10}{10^n} \cdot \frac{(n+1)4^{2n}4}{(n+2)4^{2n+2}4} \\ =\frac{10^n10}{10^n} \frac{(n+1)4^{2n}4}{(n+2)4^{2n}4^24}\]

Answer 43

you can do a lot of cancellation here

Answer 44

oh wow lol
I didn't notice the first example on that paul's note was almost this one
(basically is)

Answer 45

it's the same just no negative

Answer 46

write but the absolute values take care of the negative part )

Answer 47

one more?

Answer 48

ok

Answer 49

\[\sum (-1)^{n} \frac{ 2n }{ n+4 }\]

Answer 50

my book syas it is convergent but I don't see it because based on the alternative series test b_n the limit does not equal zero and it is increasing

Answer 51

hey question
what is the number it starts at?

Answer 52

like does it start at n=0
or 1? or something else?

Answer 53

reason I'm asking is because I want to try integral test for the bn part

Answer 54

n=1

Answer 55

the first term is -2/5

Answer 56

\[\int\limits_{\infty}^{1}\frac{2x}{x+4} dx \\ \text{ \let } u=x+4 \\ du=dx \\ \text{ so if } u=x+4 \text{ then } u-4=x\]

Answer 57

lol 1 to infinity

Answer 58

I get the same as you that it diverges

Answer 59

it looks like wolfram agrees with us too
http://www.wolframalpha.com/input/?i=sum%28%28-1%29%5En*2n%2F%28n%2B4%29%2Cn%3D1..infty%29

Answer 60

sorry it's suppose to be \[\sum_{1}^{\infty} (-1)^{n-1} \frac{ 2n }{ n+4 }\]

Answer 61

it is suppose to come from the series -2/5+4/6-6/7+8/8-10/9+...

Answer 62

your first sum is right then

Answer 63

but either way b_n diverges

Answer 64

by integral test

Answer 65

well that the series of b_n

Answer 66

alright Ill ask my teacher then the book might just be wrong. I thank you for your help

Answer 67

\[\sum_{i=1}^{\infty}b_n \text{ diverges by integral test }\]

Answer 68

np