Question

Are these series absolutely convergent, conditionally convergent, or divergent?

Answer 1

\[\sum_{n=1}^{\infty} \frac{ (-1)^{n-1} }{ 4n-5 }\]

Answer 2

I think we could use the alternating series test for this one.

Answer 3

Yes that will work!

Answer 4

Okay awesome!

Answer 5

I think it might be conditionally convergent... since b_1 = -1, b_2 = 1/3, and b_3 = 1/7

Answer 6

b_n+1 <= b_n is not fully satisfied.

Answer 7

Would you agree?

Answer 8

it is indeed conditionally convergent, but your reasoning is wrong.
"b_n+1 <= b_n is not fully satisfied" has nothing to do with conditional convergence

Answer 9

Oh... oops

Answer 10

Could you show your complete work if psble

Answer 11

The Alternating Series Test shows that a series must satisfy (1) b_n+1 >= b_n for all n (2) lim as n approaches infinity =0. b_n = 1/ (4n -5) in this case.

Answer 12

b_1 = -1, b_2 = 1/3, b_3= 1/7. Thus the series does not completely satisfy b_n+1 >= b_n

Answer 13

However it does satisfy the second part of the Alternating Series Test... lim as n approaches infinity of 1/4n -5 is zero.

Answer 14

We note that the series is convergent. Now we can go to another test to see if it is absolutely convergent.

Answer 15

Oh whoops, wrong notation. Sorry

Answer 16

I thought alternating series test requires the sequence to be "decreasing" :
b_n+1 \(\color{red}{\lt}\) b_n
right ?

Answer 17

b_ n+1 <= b_n

Answer 18

Well doesn't |b_n| converge? So wouldn't it be absolutely convergent?

Answer 19

yes

Answer 20

No, maybe lets not talk about absolute convergence yet.
If we forget about the first term, the rest of the sequence is strictly decreasing, yes ?

Answer 21

yes

Answer 22

So does the series for n > 1 meet all the requirements of alternating series test ?

Answer 23

Nope!

Answer 24

why nope ?

Answer 25

just forget about the first term and look at the remaining series

Answer 26

OH sorry, I thought you wrote n>=1. Yes

Answer 27

\[\sum_{n=2}^{\infty} \frac{ (-1)^{n-1} }{ 4n-5 }\]

So by alternating series test, this series converges.

Answer 28

Yep! Sounds good so far

Answer 29

Next see if the absolute value of the series also converges :

\[\sum_{n=2}^{\infty} \left|\frac{ (-1)^{n-1} }{ 4n-5 }\right|\]

Answer 30

thats same as
\[\sum_{n=2}^{\infty} \frac{ 1 }{ 4n-5 }\]

Answer 31

\[\frac{ 1 }{ 4n-5 }\gt \dfrac{1}{4n}\]

Since the harmonic series \(\sum\limits_{n\ge 1}\frac{1}{4n}\) diverges, the absolute value series \(\sum\limits_{n\ge 2}\frac{1}{4n-5}\) also diverges by comparison test.

Answer 32

.

Answer 33

We therefore the say the series \(\sum_{n=1}^{\infty} \frac{ (-1)^{n-1} }{ 4n-5 }\) is "conditionally convergent".

Answer 34

we say it is conditionally convergent because of below two things :
1) the actual series converges
2) the absolute value of series diverges

Answer 35

but the actual series does not converge... when n=1... it does not satisfy the alternating series test

Answer 36

forget about first term, it doesn't affect the convergence/divergence of series

Answer 37

oh since the first term is such a small term that it will not greatly affect the convergence and divergence of the series?

Answer 38

not a small term but just one term that doesn't satisfy the alternating series test.

Answer 39

you can forget about first "any finite" number of terms, they partial sum evaluates to some finite number, so they don't affect the convergence/divergence of actual series.

Answer 40

for example, you can forget about first 100 terms
you could also forget first 10^10 terms

Answer 41

because the sum evaluates to some "finite number", however big it is..

Answer 42

gotcha.

Answer 43

Okay I think I'm getting the hang of this... could you please check my work for the next two series?

Answer 44

but the word "forget first few terms" doesn't look mathematical

im just using it to make it easy for you to see what i mean..

Answer 45

Yes, I see

Answer 46

In your homework, justify your reasoning like this :

Answer 47

We know that the first finite number of the terms in the series evaluates to some finite number. Thus this does not affect the convergence or divergence of the series. We can now evaluate starting from n=2.

Answer 48

\[\sum_{n=1}^{\infty} \frac{ (-1)^{n-1} }{ 4n-5 }=-1+\color{blue}{\sum_{n=2}^{\infty} \frac{ (-1)^{n-1} }{ 4n-5 }}\]

Since the blue series converges by alternating series test,
adding a constant "-1" to the series still gives a converging series.
Therefore the overall series on left hand side also converges.

Answer 49

"We know that the first finite number of the terms in the series evaluates to some finite number"

this looks good !

Answer 50

OOHH! Wow, I remember that

Answer 51

Okay here is the next series...

Answer 52

\[\sum_{n=1}^{\infty} \frac{ 5n -1 }{ 3n +2 }\]

Answer 53

If you can help :)

Answer 54

This one is easy
try limit test

Answer 55

approaches 1

Answer 56

nope

Answer 57

5/3

Answer 58

I think I'm overthinking this xD

Answer 59

the limit of the sequence is 5/3, which is not 0.
so the series doesn't converge by limit test.

Answer 60

Would you account for the (-1)^n part when taking the limit test?

Answer 61

i don't see (-1)^n in your series ?

Answer 62

Oh sorry, I forgot it in there

Answer 63

\[\sum_{n=1}^{\infty} (-1)^{n} \frac{ 5n -1 }{ 3n +2 }\]

Answer 64

Could we use the alternating series test for this? and how?

Answer 65

when limit does the job for you, don't go for alternating series test.

Answer 66

and yes we should work the limit \(\lim\limits_{n\to\infty}(-1)^n\frac{5n-1}{3n+2}\)

Answer 67

simply working \(\lim\limits_{n\to\infty}\frac{5n-1}{3n+2}\) wont do

Answer 68

Yes agreed.

Answer 69

However that limit with (-1)^n is a bit tricky because it toggles between -1 and 1

Answer 70

Maybe using the ratio test then? I'm having difficulty choosing the right method

Answer 71

No, limit test will do.

Answer 72

Just put ur reasoning nicely

Answer 73

\[\lim\limits_{n\to\infty}(-1)^n\frac{5n-1}{3n+2}\]

Since \(\lim\limits_{n\to\infty}\frac{5n-1}{3n+2}\) evaluates to \(\frac{5}{3}\), it must be the case that \(\lim\limits_{n\to\infty}(-1)^n\frac{5n-1}{3n+2}\) toggles between \(\frac{5}{3}\) and \(-\frac{5}{3}\), which is NOT 0. Therefore the corresponding series \(\sum_{n=1}^{\infty} (-1)^{n} \frac{ 5n -1 }{ 3n +2 }\) doesn't converge by limit test.

Answer 74

That will do

Answer 75

I know this is wrong but why would the limit not be zero? would 5/3 and -5/3 cancel each other out?

Answer 76

You're confusing "limit of a sequence" with "series"

Answer 77

Oh I see now.. oops

Answer 78

\[\lim\limits_{n\to\infty}(-1)^n\frac{5n-1}{3n+2}\]

For large values of \(n\), the sequence \(a_n = (-1)^n\frac{5n-1}{3n+2}\) spits out terms that are very close to either \(\frac{5}{3}\) or \(-\frac{5}{3}\), yes ?

Answer 79

Yes.. but the limit is oscillating so it is divergent.

Answer 80

Okay I have one more series that I need help on... :)

Answer 81

Wil try, ask..

Answer 82

\[\sum_{n=1}^{\infty} \frac{ \sqrt{n-1} }{ n^{2} -6 } * (-1)^{n+1}\]

Answer 83

Could we just use the limit test like last time? So it is divergent then...

Answer 84

how do you know it diverges ?

Answer 85

Oh no, we could use the ratio test.. so that the limit of sqrt{n-1} / n^2 -6 is 0... thus it absolutely converges.

Answer 86

Yes this converges absolutely!

Answer 87

but i think your professor wants you to use alternating series test and show that the actual series converges first.


after that you can use ratio test for testing convergence of absolute value sereis

Answer 88

Alrighty! I'll make sure to do that then. Thank you ~!

Answer 89

yw:)