Question

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

Answer 1

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.

Answer 2

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<pi^2/6 so S converges.

Answer 3

* series, not geometric

Answer 4

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!

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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 \(k\)

Answer 13

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

Answer 14

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

Answer 15

Would that I could hand out more medals...

Answer 16

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

Answer 17

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 \(k\)

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

(closed form courtesy of Mathematica)

Answer 23

I thought \(\dfrac{2}{(m+1)k^2-1}\geq\dfrac{1}{n^2}\) for some reason!

\[
\min(k)=2\\
\frac{2}{4n^2-1}-\frac{1}{n^2}=\frac{2n^2-4n^2+1}{n^2(4n^2-1)}=\frac{-2n^2+1}{n^2(4n^2-1)}\leq0
\]

Answer 24

\[\sum_{m\ge0}\frac{1}{(m+1)^2k^2-1}=\sum_{m\ge1}\frac{1}{m^2k^2-1}\le\sum_{m\ge1}\frac{1}{m^2}\]
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 \(k\ge2\).)