Question

find if the series diverges or converges

Answer 1

\[\sum_{n=1}^{\infty} (-1)^n \frac{ \ln n}{ \sqrt{n} }\]

Answer 2

@rational

Answer 3

Have you tried the alternating series test?

Answer 4

yeah got an = ln n / sqrt(n) and an+1 = ln n +1 / sqrt(n+1)

Answer 5

whats the derivative of ln n / sqrt(n)

Answer 6

i know ln n is 1/n but what about sqrt(n)?

Answer 7

\[\frac{d}{dx}\left[\frac{\ln x}{\sqrt x}\right]=\frac{\sqrt x\dfrac{d}{dx}[\ln x]-\ln x\dfrac{d}{dx}[\sqrt x]}{\left(\sqrt x\right)^2}=\frac{\dfrac{\sqrt x}{x}-\dfrac{\ln x}{2\sqrt x}}{x}=x^{-3/2}-\frac{1}{2}x^{-3/2}\ln x\]

Answer 8

the statement an+1 < an is false so is it diverge?

Answer 9

No, the series does not diverge. If you want to stick to using the ratio test:
\[\lim_{n\to\infty}\left|\frac{\dfrac{(-1)^{n+1}\ln(n+1)}{\sqrt{n+1}}}{\dfrac{(-1)^n\ln n}{\sqrt n}}\right|=\lim_{n\to\infty}\frac{\sqrt n\ln(n+1)}{\sqrt{n+1}\ln n}=\lim_{n\to\infty}\frac{\ln(n+1)}{\ln n}=\cdots\]

Answer 10

Unfortunately, the ratio test is not that useful here because the limit is 1, which doesn't tell you anything about the series' behavior. I have to get going right now, but think about how to set up the alt series test.

Answer 11

okay i found that the limit equal zero \[\lim_{n \rightarrow \infty} \frac{ \ln n }{ \sqrt{n} } = \lim_{n \rightarrow \infty} \frac{ 1/n }{ 1/2\sqrt{n} }= \lim_{n \rightarrow \infty}\frac{ 2 }{ \sqrt{n} }= 0\]

Answer 12

but how do i make an + 1 < an

Answer 13

@freckles how do i make this converge?

Answer 14

yo

Answer 15

hey

Answer 16

so sith says that the series converges when an+1< an i dont understand

Answer 17

we are doing the alternating series on this one by the way

Answer 18

\[f(x)=\frac{\ln(x)}{\sqrt{x}}, x>0\]
maybe instead we can show f' is positive for x>0 ?
\[f'(x)=\frac{\frac{1}{x} \sqrt{x}-\frac{1}{2 \sqrt{x}} \ln(x)}{(\sqrt{x})^2} \\ \text{ multiplying \top and bottom by } 2x \\ f'(x)=\frac{2 \sqrt{x}-\sqrt{x} \ln(x)}{x}\]

\[f'=0 \\ 2 \sqrt{x}- \sqrt{x} \ln(x) =0 \\ \sqrt{x}(2-\ln(x))=0 \\ \text{ we have } \sqrt{x} \neq 0 \text{ since } x>0 \\ 2-\ln(x)=0 \\ \ln(x)=2 \\x=e^2\]

So hmm...
we cam test the intervals to see what is going on with f' around this number.
Like 1 would be easy to plug in and I guess 100.

What do you see?

Answer 19

I think this is shows it is increasing not decreasing so I don't think we can use the alternating test
unless I made a mistake above

Answer 20

wait sorry I don't know how to plug in numbers

Answer 21

at x=1 we have f' is positive
at x=100 we have f' is negative

Answer 22

so it is a decreasing function on (e^2,inf)

Answer 23

so if it decreases it converges

Answer 24

well yeah you need b_n>=0 for all n
and you need lim n->infty b_n=0
and you need b_n to be decreasing

to say by the alternating series test the series converge s

Answer 25

so you dont compare the an+1 and an here?

Answer 26

you try to do that

Answer 27

i did and its false

Answer 28

showing b_(n+1)<b_n for all n>=1
shows b_n is a decreasing sequence

Answer 29

I'm not sure about the alternating test
because the function isn't decreasing for all n

Answer 30

i still dont get how you know if its decreaseing or increasing

Answer 31

if f'>0 then f is increasing
if f'<0 then f is decreasing

Answer 32

oh okay

Answer 33

I sued the quotient rule above

Answer 34

found any critical numbers

Answer 35

the only critical numbers on the function'd domain was e^2
I check what happened before e^2 and checked what happened after e^2

Answer 36

I knew 1 was less than e^2
and 100 was more than e^2
so I chose those number to see what was happening on the intervals separated by x=e^2

Answer 37

why you put \[\sqrt{x}\neq 0\]

Answer 38

because it is impossible for sqrt(x) to be 0
since x is never 0

Answer 39

x>0
we didn't have x>=0

Answer 40

x equals 1 right?

Answer 41

are you talking about critical numbers?

Answer 42

or something else?

Answer 43

oh nvm i see you can plug any number for x

Answer 44

to check it's sign for f'?
well just as long as you choose a number in the interval (0,e^2)
and also a number from (e^2,infty)

Answer 45

and actually I think we can use the alternating test
since the function is only increasing for a finite amount of integers

Answer 46

e^2=7.389

\[\sum_{n=1}^{8 }(-1)^n \frac{\ln(n)}{\sqrt{n}}+\sum_{n=9}^{\infty}(-1)^n \frac{\ln(n)}{\sqrt{n}}\]
first sum is a finite number
and then adding a finite number to a converging series still gives you a converging series

Answer 47

http://tutorial.math.lamar.edu/problems/calcii/alternatingseries.aspx

Answer 48

@freckles i am doing solution number four and i dont get what they mean by the function will increase 0<x<3 and decrease x>3

Answer 49

its under step 3 by the way and another one is the last problem it saying that the function will always be decreasing for x>4

Answer 50

how does it get that?