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anonymous
 one year ago
Does the series
\[\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+1\}}\]converge?
anonymous
 one year ago
Does the series \[\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+1\}}\]converge?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Using the fact that \(\text{lcm}\{n,n+1\}=\dfrac{n(n+1)}{\text{gcd}\{n,n+1\}}\), I'm almost tempted to say that \[\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+1\}}\sim\sum_{n\ge1}\frac{1}{n^2}\] but I don't think I can jump to that conclusion just yet.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3We agree that lcm(n,n+1) is actually n(n+1). (The gcd is 1, by Euclid's algorithm) Since S=\(\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+1\}}=\sum_{n\ge1}\frac{1}{n(n+1)}<\sum_{n\ge1}\frac{1}{n^2}\) The strict inequality holds for each and every term, and since \(\sum_{n\ge1}\frac{1}{n^2}\) is a geometric series that is known to converge (=pi^2/6 prove it), so by comparison, S<pi^2/6 so S converges.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3* series, not geometric

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, since \(n\) and \(n+1\) are consecutive, that makes their lcm very easy to determine. And we can find the value of the series while we're at it quite easily: \[\sum_{n\ge1}\frac{1}{n(n+1)}=\sum_{n\ge1}\left(\frac{1}{n}\frac{1}{n+1}\right)=1\]Neat!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Nice! any two consecutive integers are coprime, gcd(n, n+1) = gcd(n, 1) = 1 so lcm(n, n+1) = n(n+1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider that the least common multiple of \(n,n+1\) is obviously \(n(n+1)\) since \(n+1n=1\) and thus they must be coprime. this reduces to $$\sum_{n=1}^\infty\frac1{n(n+1)}=\sum_{n=1}^\infty\frac{n+1n}{n(n+1)}=\sum_{n=1}^\infty\left(\frac{n+1}{n(n+1)}\frac{n}{n(n+1)}\right)\\\quad =\sum_{n=1}^\infty\left(\frac1n\frac1{n+1}\right)$$ which is obviously telescoping and since \(1/(n+1)\to0\) it converges

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0well i read that a series that converges has got some bounds and that it approaches a specific value. so in this case its going like 1/2 , 1/6, 1/12 , 1/20..... so u mark a line of length 1 on a number line and u plot the point 1/2 u knw 1/2 is still left between 1/2 and 1 and then u add 1/6 still 1/3 is left so u have bounds and u knw that it can never go > 1 so yea it converges... https://en.wikipedia.org/wiki/Convergent_series

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Could we make a more general claim here? Numerically, it would seem that \[\sum_{n\ge1}\frac{1}{\text{lcm}\{n,n+k\}}\]also converges to \(1\) for \(k\in\mathbb{N}\), though much more slowly.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for prime \(k\) the problem is barely any harder  you merely have to be careful about the multiples of \(k\): $$\sum_{n=1}^\infty\frac1{[n,n+k]}=\sum_{m=1}^\infty\sum_{n=mk+1}^{(m+1)k}\frac1{[n,n+k]}\\\quad=\sum_{m=1}^\infty\left(\sum_{n=mk+1}^{mk+k1}\frac1{n(n+k)}+\frac1{(m+1)(m+2)k}\right)$$

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[\gcd (n,n+k) \le k \]\[\Rightarrow {\rm lcm }(n,n+k) = \frac{n(n+k)}{\gcd(n,n+k)} \ge \frac{n(n+k)}{k}\]\[\Rightarrow \frac{1}{{\rm lcm} (n,n+k) }\le \frac{k}{n(n+k)}\]We can sum the rightside telescopically so I guess it is true that the sum is convergent?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, showing its convergence along those lines is exactly what I figured

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the sum of the terms for \(n\) coprime to \(k\) is accomplished with relatively little effort $$\sum_{n=mk+1}^{(m+1)k1}\frac1{n(n+k)}=\frac1k\left(\frac1{mk+1}\frac1{(m+2)k1}\right)$$ so we get: $$\frac1k\sum_{m=0}^\infty\left(\frac1{mk+1}\frac1{(m+2)k1}+\frac1{(m+1)(m+2)}\right)$$now split it up and telescope, I suppose? $$\sum\left(\frac1{mk+1}\frac1{(m+2)k1}\right)=1+\frac1{k+1}\\\sum\left(\frac1{m+1}\frac1{m+2}\right)=1$$ so we get $$\frac1k\left(2+\frac1{k+1}\right)=\frac2k+\frac1{k(k+1)}=\frac3k\frac1{k+1}$$ for prime \(k\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, ignore that  I read \((m+2)k+1\) rather than \((m+2)k1\), so that series does not quite telescope so cleanly

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0our problem term for prime \(k\) is this: $$\frac1{mk+1}\frac1{(m+2)k1}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would that I could hand out more medals...

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0How do you show \(\gcd (n,n+k) \leq k\) without using Bézout's Identity?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i mean, if (positive) \(d\) divides (positive) \(n\) and \(n+k\) then it also divides their difference, so \(d\) must divide \(k\). it follows that the greatest common divisor is at most \(k\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you don't necessarily need the full strength of Bezout's identity, just the fact that divisibility is preserved by sums which is very elementary

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\(\gcd(n,n+k)=\gcd(n.k)\le k\)

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0\[ \begin{align*} \sum_{n=1}^\infty\frac1{[n,n+k]}&=\sum_{m=0}^\infty\sum_{n=mk+1}^{(m+1)k}\frac1{[n,n+k]}\\ &=\sum_{m=0}^\infty\left(\sum_{n=mk+1}^{mk+k1}\frac1{n(n+k)}+\frac1{(m+1)(m+2)k}\right)\\ \end{align*} \] \[ \begin{align*} &\phantom{{}={}}\sum_{m=0}^\infty\sum_{n=mk+1}^{mk+k1}\frac1{n(n+k)}\\ &=\sum_{m=0}^\infty\left(\frac{1}{mk+1}\frac{1}{(m+1)k1}\right)\\ &=\sum_{m=0}^\infty\left(\frac{1}{mk+1}\frac{1}{(m+1)k+1}+\frac{1}{(m+1)k+1}\frac{1}{(m+1)k1}\right)\\ &=\sum_{m=0}^\infty\left(\frac{1}{mk+1}\frac{1}{(m+1)k+1}\frac{2}{(m+1)^2k^21}\right)\\ &=1\sum_{m=0}^\infty\frac{2}{(m+1)^2k^21} \end{align*} \] I think the sum converges but I couldn't prove it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Comparing to the series \(\sum\frac{1}{n^2}\) would suffice for convergence. It has a somewhat daunting closed form: \[\sum_{m=0}^\infty \frac{2}{(m+1)^2k^21}=\frac{k\pi\cot\dfrac{\pi}{k}}{2k}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(closed form courtesy of Mathematica)

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0I thought \(\dfrac{2}{(m+1)k^21}\geq\dfrac{1}{n^2}\) for some reason! \[ \min(k)=2\\ \frac{2}{4n^21}\frac{1}{n^2}=\frac{2n^24n^2+1}{n^2(4n^21)}=\frac{2n^2+1}{n^2(4n^21)}\leq0 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{m\ge0}\frac{1}{(m+1)^2k^21}=\sum_{m\ge1}\frac{1}{m^2k^21}\le\sum_{m\ge1}\frac{1}{m^2}\] since \[1\le k^21~~\implies~~ m^2\le m^2k^21~~\implies~~\frac{1}{m^2k^21}\le\frac{1}{m^2}\](The first inequality is true since \(k\ge2\).)
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