Question

Determine whether the following series coverges or diverges

Answer 1

Ratio test:
\[\large\lim_{n\to\infty}\frac{\frac{\ln(n+1)}{(n+1)^2}}{\frac{\ln(n)}{n^2}}\\
\lim_{n\to\infty}\frac{\ln(n+1)}{\ln(n)}\cdot\frac{n^2}{(n+1)^2}\\
\lim_{n\to\infty}\frac{\ln(n+1)}{\ln(n)}\cdot\lim_{n\to\infty}\frac{n^2}{(n+1)^2}\]

Answer 2

@SithsAndGiggles Why did you go n+1 on the first step? Unless I am misunderstanding the ratio test.

Answer 3

That's part of the ratio test.
\[\sum a_n \text{ converges if }\lim_{n\to\infty}\frac{a_{n+1}}{a_n}<1\]

Answer 4

Right. Thanks! Can I message you if I have any questions?

Answer 5

Actually, the limit is 1, so the ratio test doesn't work... I'll see what else can be done here.

Answer 6

Right. It wont works if the limit goes to 0 right?

Answer 7

I can't use the integral test because there is no anti derivative of this.

Answer 8

It will only work***

Answer 9

What about the comparison test?

Answer 10

The ratio test doesn't work if the limit is equal to 1. If the limit is greater than 1, the series diverges. If the limit is less than 1, the series converges.

I'm not sure about the comparison test just yet only because I haven't come up with a series to compare it to.

Answer 11

No wait that won't work...

Answer 12

You can try the limit comparison test, if you know it?

Answer 13

What about the alternating series test then?

Answer 14

You don't have an alternating series, so that doesn't work. I'm pretty sure the limit comparison test will though.

Answer 15

We haven't learnt the root test :/ .

Answer 16

Really? But the limit is non zero.

Answer 17

I thought because of that (-1)^n it alternates. My mistake.

Answer 18

What (-1)^n? You might have left that out.

Answer 19

Whoops!

Answer 20

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

Answer 21

Terrible sorry about that.

Answer 22

In that case, alternating series test it is.

Answer 23

Wanna help me through? I am not too familiar with alternating series.

Answer 24

\[\sum a_n\text{ converges if }\{a_n\}\text{ is a strictly decreasing sequence, and if}\\
\lim_{n\to\infty}a_n=0\]

Answer 25

Right. Thanks a lot!

Answer 26

So I end up getting it is convergent?

Answer 27

Because the limit clearly goes to 0 (albeit slowly) . And the function decreases because it's a rational function.

Answer 28

Well not a rational function but a fraction.

Answer 29

Minor correction:\[\sum a_n (-1)^n...\]
Yes, I think so. Do you have to check for absolute convergence or just for conv/divergence?

Answer 30

Just for conv/diver.

Answer 31

Let's see Sami's explanation.

Answer 32

ok i will rather use integral test
\[\Large \int\limits_{1}^{\infty}\frac{\ln}{n^2}dn=\frac{-\ln(n)}{n}|_{1}^{\infty}+\int\limits_{1}^{\infty}\frac{\ln(n)}{n}dn\]

Answer 33

Wrong sum XD .

Answer 34

I made a mistake :P .

Answer 35

You can't evaluate the integral of the sum that I put.

Answer 36

was the question ln(n)/n^2 ????
did i misread ??

Answer 37

Look below :) .

Answer 38

It was corrected later on.

Answer 39

@SithsAndGiggles : My textbook has a different explantion.

Answer 40

its alternating series . use alternating series test.

Answer 41

If \[\sum_{n=1}^{\infty}(-1)^{n-1}bn = b1 - b2 + b3 ... bn>0\]

satisfies

i) \[ b(n+1)\ge bn\] for all n

and

\[\lim_{n \rightarrow \infty}bn = 0\]

The series converges.

Answer 42

Which differs from your explanation.

Answer 43

I don't see the difference between what I said and what your book says. The first condition is another way of saying non-increasing (which is basically the same as strictly decreasing), and second is that the limit of the sequence as n approaches infinity is zero.

Answer 44

\[b_{n+1}\le b_n\]
means the next term in the sequence is less than or equal to the previous term.

Answer 45

Ohh I see now. I had a stupid moment >.< .

Answer 46

Thanks both of you :) .

Answer 47

You're welcome

Answer 48

Wait. I don't have the form (-1)^(n-1) on my equation though.

Answer 49

@SithsAndGiggles

Answer 50

I don't think the exponent on the -1 matters.
\[\sum(-1)^n\text{ and }\sum(-1)^{n-1} \text{ are both alternating series.}\]
Besides, you can always adjust the index (the starting number under sigma) in order to get the form you want.

Answer 51

How could I get the form I wanted by simply changing the number under sigma? :P .

Answer 52

Sorry for being so stubborn.

Answer 53

As an example (this isn't an alternating series, but still works),
\[\color{red}{\sum_{n=0}^\infty ar^n}=ar^0+\sum_{n=1}^\infty ar^n=\color{blue}{\sum_{n=1}^\infty ar^{n-1}}\]
Can you see why the red and blue series are the same?

Answer 54

Woudn't it be n+1 not n-1?

Answer 55

\[\begin{align*}\sum_{n=0}^\infty ar^n&=ar^0+ar^1+\cdots\\
\sum_{n=1}^\infty ar^{n-1}&=ar^{1-1}+ar^{2-1}+\cdots\\
&=ar^0+ar^1+\cdots\\
&=\sum_{n=0}^\infty ar^n\end{align*}\]
The exponent would be n+1 if you started your series at n = -1.

Answer 56

I see!

Answer 57

So :

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


Should be the same as

\[\sum_{n=3}^{\infty}\frac{ (-1)^{n-1} }{ n \ln n }\]

Answer 58

I think?

Answer 59

I forgot the change the demoninator but yeah.

Answer 60

Not quite. The n's in the denominator should also be changes to n + 1. Sorry for the confusion; I could have used a better example I think.

Answer 61

The denominator should also have n-1 instead of n.

Answer 62

N+1 ***

Answer 63

But no your explanation was fine :) . Thank you so much :) .

Answer 64

*n-1*
Sorry about that, you were right first time.

Answer 65

n-1 yeah right. >.< .

Answer 66

You're welcome!