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

1. 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.

2. 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<pi^2/6 so S converges.

3. mathmate

* series, not geometric

4. 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!

5. ganeshie8

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

6. 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

7. 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

8. 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.

9. anonymous

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

10. 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?

11. anonymous

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

12. 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 $$k$$

13. 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

14. anonymous

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

15. anonymous

Would that I could hand out more medals...

16. thomas5267

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

17. 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 $$k$$

18. 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

19. ganeshie8

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

20. 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.

21. anonymous

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

22. anonymous

(closed form courtesy of Mathematica)

23. thomas5267

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

24. 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 $$k\ge2$$.)