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- anonymous

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

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- anonymous

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

- chestercat

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- anonymous

Using 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

We 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

- mathmate

* series, not geometric

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- anonymous

Right, 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

Nice! any two consecutive integers are coprime,
gcd(n, n+1) = gcd(n, 1) = 1
so lcm(n, n+1) = n(n+1)

- anonymous

consider that the least common multiple of \(n,n+1\) is obviously \(n(n+1)\) since \(n+1-n=1\) and thus they must be coprime. this reduces to $$\sum_{n=1}^\infty\frac1{n(n+1)}=\sum_{n=1}^\infty\frac{n+1-n}{n(n+1)}=\sum_{n=1}^\infty\left(\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}\right)\\\quad =\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}\right)$$ which is obviously telescoping and since \(1/(n+1)\to0\) it converges

- imqwerty

well 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

Could 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

for prime \(k\) the problem is barely any harder -- you merely have to be careful about the multiples of \(k\): $$\sum_{n=1}^\infty\frac1{[n,n+k]}=\sum_{m=1}^\infty\sum_{n=mk+1}^{(m+1)k}\frac1{[n,n+k]}\\\quad=\sum_{m=1}^\infty\left(\sum_{n=mk+1}^{mk+k-1}\frac1{n(n+k)}+\frac1{(m+1)(m+2)k}\right)$$

- ParthKohli

\[\gcd (n,n+k) \le k \]\[\Rightarrow {\rm lcm }(n,n+k) = \frac{n(n+k)}{\gcd(n,n+k)} \ge \frac{n(n+k)}{k}\]\[\Rightarrow \frac{1}{{\rm lcm} (n,n+k) }\le \frac{k}{n(n+k)}\]We can sum the right-side telescopically so I guess it is true that the sum is convergent?

- anonymous

yeah, showing its convergence along those lines is exactly what I figured

- anonymous

the sum of the terms for \(n\) coprime to \(k\) is accomplished with relatively little effort $$\sum_{n=mk+1}^{(m+1)k-1}\frac1{n(n+k)}=\frac1k\left(\frac1{mk+1}-\frac1{(m+2)k-1}\right)$$ so we get: $$\frac1k\sum_{m=0}^\infty\left(\frac1{mk+1}-\frac1{(m+2)k-1}+\frac1{(m+1)(m+2)}\right)$$now split it up and telescope, I suppose? $$\sum\left(\frac1{mk+1}-\frac1{(m+2)k-1}\right)=1+\frac1{k+1}\\\sum\left(\frac1{m+1}-\frac1{m+2}\right)=1$$ so we get $$\frac1k\left(2+\frac1{k+1}\right)=\frac2k+\frac1{k(k+1)}=\frac3k-\frac1{k+1}$$ for prime \(k\)

- anonymous

oops, ignore that -- I read \((m+2)k+1\) rather than \((m+2)k-1\), so that series does not quite telescope so cleanly

- anonymous

our problem term for prime \(k\) is this: $$\frac1{mk+1}-\frac1{(m+2)k-1}$$

- anonymous

Would that I could hand out more medals...

- thomas5267

How do you show \(\gcd (n,n+k) \leq k\) without using Bézout's Identity?

- anonymous

i 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

you 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

\(\gcd(n,n+k)=\gcd(n.k)\le k\)

- thomas5267

\[
\begin{align*}
\sum_{n=1}^\infty\frac1{[n,n+k]}&=\sum_{m=0}^\infty\sum_{n=mk+1}^{(m+1)k}\frac1{[n,n+k]}\\
&=\sum_{m=0}^\infty\left(\sum_{n=mk+1}^{mk+k-1}\frac1{n(n+k)}+\frac1{(m+1)(m+2)k}\right)\\
\end{align*}
\]
\[
\begin{align*}
&\phantom{{}={}}\sum_{m=0}^\infty\sum_{n=mk+1}^{mk+k-1}\frac1{n(n+k)}\\
&=\sum_{m=0}^\infty\left(\frac{1}{mk+1}-\frac{1}{(m+1)k-1}\right)\\
&=\sum_{m=0}^\infty\left(\frac{1}{mk+1}-\frac{1}{(m+1)k+1}+\frac{1}{(m+1)k+1}-\frac{1}{(m+1)k-1}\right)\\
&=\sum_{m=0}^\infty\left(\frac{1}{mk+1}-\frac{1}{(m+1)k+1}-\frac{2}{(m+1)^2k^2-1}\right)\\
&=1-\sum_{m=0}^\infty\frac{2}{(m+1)^2k^2-1}
\end{align*}
\]
I think the sum converges but I couldn't prove it.

- anonymous

Comparing to the series \(\sum\frac{1}{n^2}\) would suffice for convergence. It has a somewhat daunting closed form:
\[\sum_{m=0}^\infty \frac{2}{(m+1)^2k^2-1}=\frac{k-\pi\cot\dfrac{\pi}{k}}{2k}\]

- anonymous

(closed form courtesy of Mathematica)

- thomas5267

I thought \(\dfrac{2}{(m+1)k^2-1}\geq\dfrac{1}{n^2}\) for some reason!
\[
\min(k)=2\\
\frac{2}{4n^2-1}-\frac{1}{n^2}=\frac{2n^2-4n^2+1}{n^2(4n^2-1)}=\frac{-2n^2+1}{n^2(4n^2-1)}\leq0
\]

- anonymous

\[\sum_{m\ge0}\frac{1}{(m+1)^2k^2-1}=\sum_{m\ge1}\frac{1}{m^2k^2-1}\le\sum_{m\ge1}\frac{1}{m^2}\]
since
\[1\le k^2-1~~\implies~~ m^2\le m^2k^2-1~~\implies~~\frac{1}{m^2k^2-1}\le\frac{1}{m^2}\](The first inequality is true since \(k\ge2\).)

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