Let $a_n$ such that the sequence $\sum_{n=1}^\infty a_n$ converge conditionally, does $\sum_{n=1}^\infty|a_n+a_{n+1}|$ must converge as well?
#Convergence
silver basalt
frank flareBOT
Let $a_n$ such that the sequence $\sum_{n=1}^\infty a_n$ converge conditionally,...
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oblique tartanBOT
mtr123
silver basalt
my intuition is that the claim is wrong, though trying to come up with a counter example wasn't easy for me, tried (-1)^n/n but that doesn't work
warped kraken
if alternating one term doesn't work, ||how about try alternating two terms?||
silver basalt
if alternating one term doesn't work, ||how about try alternating two terms?||
yeah ill give it a shot
silver basalt
if alternating one term doesn't work, ||how about try alternating two terms?||
Ok i think i got it
choose:
$a_n = \frac{(-1)^n}{ln(n)}$ we have $\sum_{n=1}^\infty 1/ln(n) \geq \sum_{n=1}^\infty 1/n$ diverge and also by leibniz theorem we have $\sum_{n=1}^\infty a_n$ converge
now we check for $a_n+a_{n+1}$
$= | (-1)^n \frac{1}{ln(n) - \frac{1}{ln(n+1)}} | 😐 \frac{ln(n+1)-ln(n)}{ln(n)ln(n+1)}|$
now i want to compare this with $\frac{1}{ln(n)^2}$ and claim that they diverge or converge together and i know this series diverge since its bigger than 1/(sqrt n)^2 which diverge and thats it?
oblique tartanBOT
mtr123
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stone orbit
Ok i think i got it
choose:
$a_n = \frac{(-1)^n}{ln(n)}$ we have $\sum_{n=1}^\...
Unfortunately you'll find that the series converges
the denominator is equivalent to ln(n)²
but try finding an equivalent of the numerator
silver basalt
which part are you relating too?
ain't sum a_n+a_n+1 diverge?
silver basalt
dont i get ln(n+1 / n) and as goes to infty it goes to 0 cuz we have ln(1)?
so they diverge/ converge together?
stone orbit
what's an equivalent of that?
stone orbit
so they diverge/ converge together?
yes but it doesn't diverge
silver basalt
1/ln(n)^2 converge?
stone orbit
but that's not an equivalent of your general term
silver basalt
dont we have ln(n) < sqrt n?
so 1/ln(n)^2 > 1 / n?
stone orbit
your general term is smaller
stone orbit
yes but that isn't enough to conclude that |a_n + a_{n+1}| has a divergent series
because |a_n + a_[n+1]| in practice is much smaller than 1/ln(n)²
silver basalt
can't we use this one?
with c_n at the numerator = |a_n+a_{n+1}|
and b_n = 1/ln(n)^2?
this is what im suggesting but maybe im missing omething
stone orbit
yes but you'll find out that their ratio isn't convergent to a nonzero finite limit
in other words they're not equivalent
$\frac{\ln(1 + \frac{1}{n})}{\ln(n)\ln(n+1)}$
oblique tartanBOT
Rion
silver basalt
and we divide that by 1/ln(n)^2
so we are left with $\frac{ln(n)}{ln(n+1)}\cdot ln(1+\frac{1}{n})$
oblique tartanBOT
mtr123
stone orbit
yeah see, now that clearly doesn't converge to a finite nonzero limit
stone orbit
which is 0
that's not nonzero
stone orbit
in fact $\lvert a_n - a_{n+1} \rvert$ is equivalent to $\frac{1}{n \ln n}$
oblique tartanBOT
Rion
silver basalt
so this example doesn't even work?
stone orbit
yes
try following MartinFTW's suggestion
take $a_{2n} = a_{2n+1} = \frac{(-1)^n}{n}$
oblique tartanBOT
Rion
silver basalt
you wrote is as 2n and 2n+1 for even / odd?
stone orbit
yeah
take a_0 and a_1 some arbitrary value
(it doesn't matter)
a_2 = a_3
a_4 = a_5
...
- why is that the term of a conditionally convergent series?
- verify the other property
silver basalt
ain't it actually:
$a_2n = 2\cdot \frac{(-1)^n}{n}?$
so it will diverge easily if we take absolute value and converge again using Leibniz.
need to check the last property
oblique tartanBOT
mtr123
stone orbit
ain't it actually:
$a_2n = 2\cdot \frac{(-1)^n}{n}?$
so it will diverge easily ...
what?
I don't get what the 2 factor is doing there
silver basalt
summing the original series, ain't it like looking at the even spots and multiply by 2? cuz odd=even?
a_2=a_3
a_4=a_5
so we can just sum the even spots and multiply by 2, cant we?
stone orbit
basically, I took a_2 = 1/2, a_3 = 1/2, a_4 = -1/3, a_5 = -1/3, ...
stone orbit
summing the original series, ain't it like looking at the even spots and multipl...
What you mean is $\sum a_n = 2 \sum \frac{(-1)^n}{n}$
oblique tartanBOT
Rion
stone orbit
this much is true, but needs to be proven (fall back to partial sums)
silver basalt
ok i see, let me think about it a little bit and i'll just write here later on
silver basalt
ok so for the first part, can't i still you Leibniz theorem to show this series indeed converge?
in order to show it doens't absolutely converge we can look at
$\sum_{n=1}^\infty a_{2n} = \frac{1}{n}$ which still diverge cuz its just the harmonic series and it is less than the original absolute value of the series.
oblique tartanBOT
mtr123
silver basalt
and for the last part we have
$a_{2n}+a_{2n+1} = \frac{2}{n}$ so since all our sequence values are positive (taking absolute value), and this part diverge, the series is going to diverge?
oblique tartanBOT
mtr123
stone orbit
@silver basalt
Formally, to show convergence, consider $S_{N} = \sum_{n=0}^{N} a_n = \sum_{ 2 \leq n \leq N \mathrm{ even}} a_n + \sum_{ 2 \leq n \leq N \mathrm{ odd}} a_n$
oblique tartanBOT
Rion
stone orbit
now notice that while both these sums might not be equal, they both converge
and this is where you start to think of Leibniz's theorem to show it
silver basalt
oh so since both converge i can claim the whole series converge so i could use it?
stone orbit
oh so since both converge i can claim the whole series converge so i could use i...
Why do both converge?
silver basalt
dont we just get the lebiniz series for each?
$\sum_{n=1}^\infty (-1)^n/n$
oblique tartanBOT
mtr123
stone orbit
dont we just get the lebiniz series for each?
$\sum_{n=1}^\infty (-1)^n/n$
In case you don't get why I'm asking so many questions: the task is quite reasoning intensive and I need you to be quite rigorous in the reasoning, which includes using the right terms too
You can indeed apply Leibniz's rule to each series to prove they converge
hence, the partial sum S_N will also converge as N tends to infinity
silver basalt
yeah no problem, i like that you keep asking questions cuz it helps
silver basalt
hence, the partial sum S_N will also converge as N tends to infinity
cool thank you!
stone orbit
Now that you've shown that the series of a_n converges (because the partial sum converges)
can you disprove absolute convergence?
again, here, use partial sums to formally disprove absolute convergence
silver basalt
actually not sure how to use partial sums here.
looking at the even n's we get $\sum_{n=1}^\infty = \frac{1}{n}$ and the same for the odd part. meaning both are the harmonic sequence which sum diverge to infinity so can't we just claim it diverge?
oblique tartanBOT
mtr123
silver basalt
because we have just even is smaller than even + odd (all are positive values)
stone orbit
actually not sure how to use partial sums here.
looking at the even n's we get ...
Yes that's the idea
you can find a divergent subsequence of (A_N), where A_N is the partial sum of |a_n| up to the N-th term
take (A_2N), it is easy to see that it diverges
therefore S_N does not converge
stone orbit
sorry for the delay
Don't worry, all that's left is to conclude
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