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Loser66
 one year ago
\(\sum_{n =1}^\infty (1/n) \) converges. How about \(\sum_{n=1}^\infty (1/n)\) ? converges also? Please, help
Loser66
 one year ago
\(\sum_{n =1}^\infty (1/n) \) converges. How about \(\sum_{n=1}^\infty (1/n)\) ? converges also? Please, help

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idku
 one year ago
Best ResponseYou've already chosen the best response.0sum from 1 to ∞ of 1/n Hormonic series

idku
 one year ago
Best ResponseYou've already chosen the best response.0Converges means sum is defined

idku
 one year ago
Best ResponseYou've already chosen the best response.0diverges means the sum doesn't exist.

idku
 one year ago
Best ResponseYou've already chosen the best response.0In case you mixed up terms

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \displaystyle\sum_{n=1}^{\infty}\frac{1}{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...\) this is at least: \(\large \displaystyle\sum_{n=1}^{\infty}\frac{1}{n}\ge\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+...\) \(\large \displaystyle\sum_{n=1}^{\infty}\frac{1}{n}\ge\frac{1}{1}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+...\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0ok, I think I mess up something. Sorry about that .

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Harmonic 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 \(\ln 2\)!

idku
 one year ago
Best ResponseYou've already chosen the best response.0alternating series test. Converges

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Let me post the original one.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0\(\sum_{n=1}^\infty \dfrac{(1)^n i^{n(n+1)}}{n}\) converge or diverge?

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \displaystyle\sum_{n=1}^{\infty}\frac{(1)^n}{n}\) it meets conditions: \(\large \displaystyle \lim_{k\to \infty} (a_n)=0\) \(\large \displaystylea_n>a_{n+1}\) for all n. Thus, converges.

idku
 one year ago
Best ResponseYou've already chosen the best response.0i mean in the limit "n > "

idku
 one year ago
Best ResponseYou've already chosen the best response.0what does i do in your series?

idku
 one year ago
Best ResponseYou've already chosen the best response.0is that *i* to indicate imaginary value, or what>?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I lost the net......:(

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Please, leave your guide here, I will take it if I can see what you guys write.

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \displaystyle\sum_{n=1}^{\infty}\frac{(1)^ni^{n(n1)}}{n}\) I haven't done series with imaginary numbers mixed in...... my guess is: \(\large \displaystyle\sum_{n=1}^{\infty}\frac{(1)^ni^{n(n1)}}{n}\le\large \displaystyle\sum_{n=1}^{\infty}\frac{(1)^n}{n}=\rm converges\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[n=1\Rightarrow i^{n(n+1)}=1\]\[n=2 \Rightarrow i^{n(n+1)}=1\]\[n=3 \Rightarrow i^{n(n+1)}=1\]\[n=4 \Rightarrow i^{n(n+1)}=1\]And this continues.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0\[a_{n}=\frac{ 1 }{ n }\] \[\lim_{x \rightarrow \infty}\frac{ 1 }{ n } =0\] so because the limit is approaching a certain value we will say that it converges

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Nah, it diverges @imqwerty

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Even though the limit of the sequence exists and is equal to zero, that is not sufficient to say that the series should converge

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0but whenever we take the limit and if it approaches certain value then it converges right?

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \displaystyle\sum_{n=1}^{\infty}\frac{(1)^ni^{n(n1)}}{n}\) \(\large \displaystyle\sum_{n=1}^{\infty}\left(\frac{(1)^{n+1}}{n}+\frac{(1)^n}{n+1}\right)\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1first of all, for convergence, the limit has to be zero and no other value. but that is not sufficient for convergence.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0we will also consider the limit of partial sums right?

idku
 one year ago
Best ResponseYou've already chosen the best response.0yes, true dirverges test only shows divergence if lim \(\ne\)0

idku
 one year ago
Best ResponseYou've already chosen the best response.0divergence test ... i meant

idku
 one year ago
Best ResponseYou've already chosen the best response.0I was thinking it is an equivalent representation: \(\large \displaystyle\sum_{n=1}^{\infty}\frac{(1)^ni^{n(n+1)}}{n}=\sum_{n=1}^{\infty}\left(\frac{(1)^{n+1}}{n}+\frac{(1)^n}{n+1}\right)\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.0The divergence test is not called the convergence test for the reason that it can only show you if it diverges or not.

idku
 one year ago
Best ResponseYou've already chosen the best response.0But is this reasonable? (What I said above)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0why cant we say that a series diverges or converges by jst calculating the limit ?

Empty
 one year ago
Best ResponseYou've already chosen the best response.0The 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.

idku
 one year ago
Best ResponseYou've already chosen the best response.0Because when the sequence an approaches 0, that can be a divergent series \(a_n) eg Harmonic series

idku
 one year ago
Best ResponseYou've already chosen the best response.0If 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

Empty
 one year ago
Best ResponseYou've already chosen the best response.0you 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.

idku
 one year ago
Best ResponseYou've already chosen the best response.0and that would in fact be: ln(2)+ [ ln(2)1 ] = 2ln(2) 1

idku
 one year ago
Best ResponseYou've already chosen the best response.0for my representation, if it is right

idku
 one year ago
Best ResponseYou've already chosen the best response.0I checked the signs, so should be fine.

idku
 one year ago
Best ResponseYou've already chosen the best response.0so, as parth said the i's yield: 1 1 1 1

Empty
 one year ago
Best ResponseYou've already chosen the best response.0I don't understand I thought the problem was \[\sum_{n=1}^\infty \frac{(1)^{n+1}}{n}\]

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \displaystyle\sum_{n=1}^{\infty}\frac{(1)^ni^{n(n+1)}}{n}\) this is the prob

idku
 one year ago
Best ResponseYou've already chosen the best response.0and then I suggest that just as: \(\displaystyle\sum_{n=1}^{\infty}\frac{(1)^ni^{n(n+1)}}{n}=1/11/21/3+1/4+1/51/61/7+1/8...\) so is: \(\displaystyle\sum_{n=1}^{\infty}\left(\frac{(1)^{n+1}}{n}+\frac{(1)^{n}}{n+1}\right)=1/11/21/3+1/4+1/51/61/7+1/8...\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.0I was looking at the wrong problem no wonder haha

idku
 one year ago
Best ResponseYou've already chosen the best response.0i was wrong, because my terms cancel, I have to skeep over 1

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\displaystyle\large\sum_{n=0}^{\infty}\left(\frac{(1)^{n}}{2n+1}+\frac{(1)^{n+1}}{2n+2}\right)\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0this 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 \(\left[\pi2\ln2\right]/2\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Actually 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: \[\Re(S) = \frac{S+S^*}{2}\]\[\Im(S) = \frac{SS^*}{i2}\] 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.

idku
 one year ago
Best ResponseYou've already chosen the best response.0Loser66, how are you so far?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0:) 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?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0for 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?

idku
 one year ago
Best ResponseYou've already chosen the best response.0I hope you aren't factoring all that imaginary component out of the series......

idku
 one year ago
Best ResponseYou've already chosen the best response.0also, it is not (1)^(n(n+1)), but (i)^(n(n+1))

idku
 one year ago
Best ResponseYou've already chosen the best response.0so it is not always positive

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Hey, all natural number can be expressed as one of the forms : 4k, 4k+1, 4k+2 , 4k+3.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Why do we have to do that? because that is the way we turn imaginary i to real since i^2 =1

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0and the circle to get real number from i is 4. I was taught that. :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0\(\sum_{n=1}^\infty a_n = 0~~ if~~ n \cong 0~~~ mod4\) \(\sum_{n=1}^\infty a_n = 1~~ if~~ n \cong 1~~~ mod4\) \(\sum_{n=1}^\infty a_n = 0~~ if~~ n \cong2~~~ mod4\) \(\sum_{n=1}^\infty a_n = 1~~ if~~ n \cong 3~~~ mod4\) That shows the sum is divergent.
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