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

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

Mathematics
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diverges!
why?

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sum from 1 to ∞ of 1/n Hormonic series
Converges means sum is defined
diverges means the sum doesn't exist.
In case you mixed up terms
\(\large \displaystyle\sum_{n=1}^{\infty}\frac{1}{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...\) this is at least: \(\large \displaystyle\sum_{n=1}^{\infty}\frac{1}{n}\ge\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+...\) \(\large \displaystyle\sum_{n=1}^{\infty}\frac{1}{n}\ge\frac{1}{1}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...\)
ok, I think I mess up something. Sorry about that .
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 \(\ln 2\)!
alternating series test. Converges
Let me post the original one.
\(\sum_{n=1}^\infty \dfrac{(-1)^n i^{n(n+1)}}{n}\) converge or diverge?
\(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\) it meets conditions: \(\large \displaystyle \lim_{k\to \infty} (a_n)=0\) \(\large \displaystyle|a_n|>|a_{n+1}\) for all n. Thus, converges.
i mean in the limit "n -> "
what does i do in your series?
is that *i* to indicate imaginary value, or what>?
it does
I lost the net......:(
Please, leave your guide here, I will take it if I can see what you guys write.
\(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n-1)}}{n}\) I haven't done series with imaginary numbers mixed in...... my guess is: \(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n-1)}}{n}\le\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=\rm converges\)
\[n=1\Rightarrow i^{n(n+1)}=-1\]\[n=2 \Rightarrow i^{n(n+1)}=-1\]\[n=3 \Rightarrow i^{n(n+1)}=1\]\[n=4 \Rightarrow i^{n(n+1)}=1\]And this continues.
\[a_{n}=\frac{ -1 }{ n }\] \[\lim_{x \rightarrow \infty}\frac{ -1 }{ n } =0\] so because the limit is approaching a certain value we will say that it converges
Nah, it diverges @imqwerty
Even though the limit of the sequence exists and is equal to zero, that is not sufficient to say that the series should converge
but whenever we take the limit and if it approaches certain value then it converges right?
nope...
\(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n-1)}}{n}\) \(\large \displaystyle\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^n}{n+1}\right)\)
is that right?
first of all, for convergence, the limit has to be zero and no other value. but that is not sufficient for convergence.
we will also consider the limit of partial sums right?
yes, true dirverges test only shows divergence if lim \(\ne\)0
divergence test ... i meant
yes
I was thinking it is an equivalent representation: \(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n+1)}}{n}=\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^n}{n+1}\right)\)
The divergence test is not called the convergence test for the reason that it can only show you if it diverges or not.
yes.
But is this reasonable? (What I said above)
why cant we say that a series diverges or converges by jst calculating the limit ?
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.
Because when the sequence an approaches 0, that can be a divergent series \(a_n|) eg Harmonic series
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
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.
and that would in fact be: ln(2)+ [ ln(2)-1 ] = 2ln(2) -1
for my representation, if it is right
I checked the signs, so should be fine.
i is imaginary
so, as parth said the i's yield: -1 -1 1 1
I don't understand I thought the problem was \[\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\]
\(\large \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^ni^{n(n+1)}}{n}\) this is the prob
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: \(\displaystyle\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^{n}}{n+1}\right)=1/1-1/2-1/3+1/4+1/5-1/6-1/7+1/8...\)
I was looking at the wrong problem no wonder haha
oh no not that
i was wrong, because my terms cancel, I have to skeep over 1
\(\displaystyle\large\sum_{n=0}^{\infty}\left(\frac{(-1)^{n}}{2n+1}+\frac{(-1)^{n+1}}{2n+2}\right)\)
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 \(\left[\pi-2\ln2\right]/2\)
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.
Loser66, how are you so far?
:) 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?
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?
I hope you aren't factoring all that imaginary component out of the series......
also, it is not (-1)^(n(n+1)), but (i)^(n(n+1))
so it is not always positive
-1, -1, 1, 1.....
Hey, all natural number can be expressed as one of the forms : 4k, 4k+1, 4k+2 , 4k+3.
So as n
Why do we have to do that? because that is the way we turn imaginary i to real since i^2 =-1
and the circle to get real number from i is 4. I was taught that. :)
\(\sum_{n=1}^\infty a_n = 0~~ if~~ n \cong 0~~~ mod4\) \(\sum_{n=1}^\infty a_n = -1~~ if~~ n \cong 1~~~ mod4\) \(\sum_{n=1}^\infty a_n = 0~~ if~~ n \cong2~~~ mod4\) \(\sum_{n=1}^\infty a_n = 1~~ if~~ n \cong 3~~~ mod4\) That shows the sum is divergent.

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