Question

\(\sum_{n =1}^\infty (1/n) \) converges. How about \(\sum_{n=1}^\infty -(1/n)\) ? converges also? Please, help

Answer 1

@imqwerty

Answer 2

diverges!

Answer 3

why?

Answer 4

sum from 1 to ∞
of 1/n

Hormonic series

Answer 5

Converges means sum is defined

Answer 6

diverges means the sum doesn't exist.

Answer 7

In case you mixed up terms

Answer 8

\(\large \displaystyle\sum_{n=1}^{\infty}\frac{1}{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...\)

this is at least:

\(\large \displaystyle\sum_{n=1}^{\infty}\frac{1}{n}\ge\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+...\)

\(\large \displaystyle\sum_{n=1}^{\infty}\frac{1}{n}\ge\frac{1}{1}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...\)

Answer 9

ok, I think I mess up something. Sorry about that .

Answer 10

Harmonic series diverges. For the second, if you mean\[\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n} = 1-\frac{1}{2} + \frac{1}3 - \frac{1}4 + \cdots\]The second sum is in fact convergent and equal to \(\ln 2\)!

Answer 11

alternating series test. Converges

Answer 12

Let me post the original one.

Answer 13

\(\sum_{n=1}^\infty \dfrac{(-1)^n i^{n(n+1)}}{n}\) converge or diverge?

Answer 14

\(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\)

it meets conditions:

\(\large \displaystyle \lim_{k\to \infty} (a_n)=0\)

\(\large \displaystyle|a_n|>|a_{n+1}\) for all n.

Thus, converges.

Answer 15

i mean in the limit "n -> "

Answer 16

what does i do in your series?

Answer 17

is that *i* to indicate imaginary value, or what>?

Answer 18

it does

Answer 19

I lost the net......:(

Answer 20

Please, leave your guide here, I will take it if I can see what you guys write.

Answer 21

\(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n-1)}}{n}\)

I haven't done series with imaginary numbers mixed in......
my guess is:

\(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n-1)}}{n}\le\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=\rm converges\)

Answer 22

\[n=1\Rightarrow i^{n(n+1)}=-1\]\[n=2 \Rightarrow i^{n(n+1)}=-1\]\[n=3 \Rightarrow i^{n(n+1)}=1\]\[n=4 \Rightarrow i^{n(n+1)}=1\]And this continues.

Answer 23

\[a_{n}=\frac{ -1 }{ n }\]
\[\lim_{x \rightarrow \infty}\frac{ -1 }{ n } =0\]
so because the limit is approaching a certain value we will say that it converges

Answer 24

Nah, it diverges @imqwerty

Answer 25

Even though the limit of the sequence exists and is equal to zero, that is not sufficient to say that the series should converge

Answer 26

but whenever we take the limit and if it approaches certain value then it converges right?

Answer 27

nope...

Answer 28

\(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n-1)}}{n}\)

\(\large \displaystyle\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^n}{n+1}\right)\)

Answer 29

is that right?

Answer 30

first of all, for convergence, the limit has to be zero and no other value. but that is not sufficient for convergence.

Answer 31

we will also consider the limit of partial sums right?

Answer 32

yes, true dirverges test only shows divergence if lim \(\ne\)0

Answer 33

divergence test ... i meant

Answer 34

yes

Answer 35

I was thinking it is an equivalent representation:

\(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n+1)}}{n}=\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^n}{n+1}\right)\)

Answer 36

The divergence test is not called the convergence test for the reason that it can only show you if it diverges or not.

Answer 37

yes.

Answer 38

But is this reasonable?
(What I said above)

Answer 39

why cant we say that a series diverges or converges by jst calculating the limit ?

Answer 40

The intuitive reason for what happens when you use the divergence test is this:

You have an infinite sum. If the 'last' term isn't 0, then you know it diverges since infinity never ends.

Answer 41

Because when the sequence an approaches 0, that can be a divergent series \(a_n|)

eg
Harmonic series

Answer 42

If I am right,

\(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n+1)}}{n}=\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^n}{n+1}\right)\)

then the series converges

Answer 43

you can have your 'last' term equal to 0 in an infinite sum, but you already went through an infinite number of terms prior to that, so those could still possibly diverge.

Answer 44

and that would in fact be: ln(2)+ [ ln(2)-1 ] = 2ln(2) -1

Answer 45

for my representation, if it is right

Answer 46

I checked the signs, so should be fine.

Answer 47

i is imaginary

Answer 48

so, as parth said the i's yield:
-1
-1
1
1

Answer 49

I don't understand I thought the problem was

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

Answer 50

\(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n+1)}}{n}\)

this is the prob

Answer 51

and then I suggest that just as:

\(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n+1)}}{n}=1/1-1/2-1/3+1/4+1/5-1/6-1/7+1/8...\)

so is:

\(\displaystyle\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^{n}}{n+1}\right)=1/1-1/2-1/3+1/4+1/5-1/6-1/7+1/8...\)

Answer 52

I was looking at the wrong problem no wonder haha

Answer 53

oh no not that

Answer 54

i was wrong, because my terms cancel, I have to skeep over 1

Answer 55

\(\displaystyle\large\sum_{n=0}^{\infty}\left(\frac{(-1)^{n}}{2n+1}+\frac{(-1)^{n+1}}{2n+2}\right)\)

Answer 56

this should be better:

=
+1/1 - 1/2 - 1/3 + 1/4
+1/5 - 1/6 - 1/7 + 1/8
+1/9 - 1/10 - 1/11 + 1/12

etc....

but that cycle is essentially just a combo of two altern. series.
\(\pi/4\) + -ln(2)/2

\(\left[\pi-2\ln2\right]/2\)

Answer 57

Actually we can probably evaluate it but to see if this series converges or diverges I think we can use a bit of a trick, look at the real and imaginary parts separately:

\[\Re(S) = \frac{S+S^*}{2}\]\[\Im(S) = \frac{S-S^*}{i2}\]

These are both going to be completely real series themselves, and you can use the alternating series test to check for convergence on both:

\[\Re(S) = \sum_{n=1}^\infty \frac{(-1)^n}{n} \cos\left( \frac{\pi}{2} n(n+1) \right) \]
\[\Im(S) = \sum_{n=1}^\infty \frac{(-1)^n}{n} \sin\left( \frac{\pi}{2} n(n+1) \right) \]

Inside you have \(\frac{\pi}{2}\) so every 4 values of n you'll repeat so you could separate it out like this into 4 terms, find out what the cosine and sine terms are doing, they'll only give you 0, 1, or -1 in some funny order at worst, but we can see that the inside will repeat every 4 because of this:
\[\cos[ \frac{\pi}{2}n(n+1)]\]
\[n \to n+4\]
\[\cos [ \frac{\pi}{2}(n+4)(n+4+1)] = \cos[ \frac{\pi}{2}n(n+1)+\frac{\pi}{2}(4n+4n+4)]=\cos[ \frac{\pi}{2}n(n+1)]\]

really it's just case \(\frac{\pi}{2}(4n+4n+4)\) is a multiple of \(2 \pi\). Maybe this ends up sounding more complicated than just solving this damn thing I don't know haha. But it's just 4 terms haha.

Answer 58

Loser66, how are you so far?

Answer 59

:) To me, I do something like
let \(a_n = 1/n\) \(b_n = (-1)^n i^{n(n+1)}\)
Now, just calculate \(b_n\)
Note: \(i^2 =-1\\n(n+1) is~~even\)
Hence, for n = 4k, \(b_n = (-1)^4k i^{4k (4k +1) }= 1* ((i^2)^2)^{k(4k+1)}=1\)
Hence \(b_n =1, \sum_{n=1}^\infty a_n b_n = \sum_{n=1}^\infty \dfrac{1}{n}\). You say, is it divergent or convergent?

Answer 60

for n= 4k+1, similar argument
\[b_n = (-1)^{4k+1}i^{(4k+1)(4k+2)}= -1* (i^2)^{(4k+1)(2k+1)}= -1*-1=1\]

Hence \[\sum_{n=1}^\infty a_n b_n = \sum_{n=1}^\infty \dfrac{1}{n}\]
You say, is it divergent or convergent?

Answer 61

I hope you aren't factoring all that imaginary component out of the series......

Answer 62

also, it is not (-1)^(n(n+1)), but (i)^(n(n+1))

Answer 63

so it is not always positive

Answer 64

-1, -1, 1, 1.....

Answer 65

Hey, all natural number can be expressed as one of the forms : 4k, 4k+1, 4k+2 , 4k+3.

Answer 66

So as n

Answer 67

Why do we have to do that? because that is the way we turn imaginary i to real since i^2 =-1

Answer 68

and the circle to get real number from i is 4. I was taught that. :)

Answer 69

\(\sum_{n=1}^\infty a_n = 0~~ if~~ n \cong 0~~~ mod4\)

\(\sum_{n=1}^\infty a_n = -1~~ if~~ n \cong 1~~~ mod4\)

\(\sum_{n=1}^\infty a_n = 0~~ if~~ n \cong2~~~ mod4\)

\(\sum_{n=1}^\infty a_n = 1~~ if~~ n \cong 3~~~ mod4\)

That shows the sum is divergent.