Question

test the series for convergence or divergence. Use comparison or limit comparison test.

Answer 1

\[\sum_{n=2}^{\infty} \frac{ \ln n }{ n }\]

Answer 2

@jim_thompson5910 could you help me please

Answer 3

i got an to be \[\frac{ \ln n }{ n } and.....bn=\frac{ 1 }{ n}\]

Answer 4

i am not sure if i am doing it right or not

Answer 5

I suggest the integral test

Answer 6

i cant do this using the comparison or limit test?

Answer 7

one moment

Answer 8

alright

Answer 9

ok u know that 1/n^p p has to be atleast greater than 1 to converge

Answer 10

yeah

Answer 11

and ln n / n > 1/n

Answer 12

yes, you can use limit comparison.

compare 1/n to ln(n)/n

\[\lim_{n \rightarrow infinity } (b _{n}/a _{n})\]

bn is 1/n
an is ln n/n

1/n * n/ln(n) = lim 1/ ln(n) = infinity, not defined. So diverge.

Answer 13

You have the correct \(\Large a_n\) and \(\Large b_n\)

Solve ln(n)/n = 1/n, to get

ln(n)/n = 1/n
ln(n) = 1
n = e^1
n = e
n = 2.718 (approximately)

Notice how if n > 2, say n = 3, then we will have

ln(n)/n = ln(3)/3 = 0.3662
1/n = 1/3 = 0.33333

So this shows us that 1/n < ln(n)/n for all n such that n > 2

You can check other values of n that are larger than 2 and you'll find 1/n < ln(n)/n holds true

Answer 14

limit comparison test is nice here as it avoids all above algebra
simply divide and take the limit

Answer 15

you can use a p-series test to find that 1/n diverges

Let
\[\Large a_{n} = \frac{\ln(n)}{n}\]
\[\Large b_{n} = \frac{1}{n}\]

Since \(\Large b_{n} < a_{n} \) for n > 2 (n is an integer), and the b sequence diverges, this means the 'a' sequence must also diverge as well

Answer 16

Either method works, so you can take your pick.

Answer 17

so what if its \[\frac{ 1 }{1+e^n }\]

Answer 18

would bn still equal 1/n?

Answer 19

whats the full question ? 1/(1+e^n) is not a series

Answer 20

its \[\sum_{n=1}^{\infty} \frac{ 1 }{ 1+e^n }\]

Answer 21

i was thinking maybe bn = \[\frac{ 1 }{ e^n } \]

Answer 22

try comparing 1/(1+e^n) with 1/n^2

Answer 23

depends on what series you already know converges

Answer 24

if you already know that SUM 1/e^n converges then you may use that for comparison

Answer 25

i prefer 1/n^2 though as it is more popular

Answer 26

You can also use the integral test

Answer 27

\[\frac{ 1 }{ 1+e^n } \ge \frac{ 1 }{ n^2 }\]

Answer 28

ou're doing it reverse

For convergence using comparison, you need to show that the terms in sequence {1/(1+a^n)} are LESS than the terms in sequence {1/n^2}

Answer 29

i can but my teacher wants us using the comparison or limit comparison test

Answer 30

for \(n\gt 0\) we have \[e^n+1 \gt n^2 \implies \dfrac{1}{e^n+1} \lt \dfrac{1}{n^2} \]

since the series \(\sum \frac{1}{n^2}\) converges,
the series \(\sum \frac{1}{e^n+1}\) also converges by direct comparison test.

Answer 31

how do you use the direct comparison test in this problem?

Answer 32

"direct comparison test" is another name for "comparison test"

Answer 33

both are same

Answer 34

oh okay and when do you know its divergence?

Answer 35

lets see a quick example :

64 + 32 + 16 + 8 + 4 + 2 + ...

what do you know about convergence/divergence of above series ?

Answer 36

the series is decreasing

Answer 37

right, thats a good reason to suspect that the series can possibly converge

Answer 38

careful when you say : "the series is decreasing"

that statement makes no sense

Answer 39

i know you meant "the terms of series are decreasing"

Answer 40

like for example this one \[\sum_{n=1}^{\infty} \frac{ 1 }{ n^n}..... diverges\]

Answer 41

because \[\frac{ 1 }{ n^n } > \frac{ 1 }{ n^2 }\]

Answer 42

right?

Answer 43

\[n^n \gt n^2 \implies \frac{1}{n^n} \lt \frac{1}{n^2}\]

so the series SUM 1/n^n actually converges

Answer 44

10 > 2

so

1/10 < 1/2

Answer 45

2 > 1

so

1/2 < 1/1

Answer 46

but look if n =3 3^3>3^2

Answer 47

right?

Answer 48

right,

3^3 > 3^2

so we have

1/3^3 < 1/3^2

Answer 49

oh yeah forgot they were fractions

Answer 50

so no need to find the limit for this one since we already know it converges right?

Answer 51

comparison test doesn't require working the limit

can we go back to earlier example series and finish it off ?

Answer 52

oh yes

Answer 53

```
64 + 32 + 16 + 8 + 4 + 2 + ...
```

what do you know about convergence/divergence of above series ?

Answer 54

well the term of the sequence is decreasing by 32/n or something like that

Answer 55

right so what can you say about its convergence ?

Answer 56

the sequence and series converges?

Answer 57

why ?

Answer 58

i dont really know

Answer 59

not covered geometric series yet ?

Answer 60

no

Answer 61

but could you tell me what you mean?

Answer 62

```
64 + 32 + 16 + 8 + 4 + 2 + ...
```
suppose we know that that series converges then the below series also converges by "comparison test" because each term in below series is positive and is less than the corresponding term in above series which converges.
```
63 + 31 + 15 + 7 + 3 + 1 + ...
```

Answer 63

i was just trying to show you why "comparison test" works

but that example wont make sense if "geometric series" is not studied yet

Answer 64

is geometric series similar to comparison test?

Answer 65

geometric series is name of an actual series

comparison test is a test for knowing whether a series converges or diverges


both are different things

Answer 66

could you show me an example of a geometric series?

Answer 67

see this
https://www.khanacademy.org/math/precalculus/seq_induction/infinite-geometric-series/v/infinite-geometric-series