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This just came out of curiosity let $$L(n)=lcm(1,2,3 \cdots n)$$ and I know that we can write this with the help of some product involving primes and all . But what I am interested is in

Does $$\sum^{\infty}_{n=1} \frac{L(n)}{(n!)^2}$$ exists ? If yes than what is it equal to ?And another thing is does $$\sum_{n=1}^{\infty} \frac{1}{L(n)}$$ exists ? If yes is there a close form for it?

Shivam Patel
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  • Since $(n-1)n \leqslant L_n \leqslant n!$, both series converge. I don't expect a nice closed form for either. – Daniel Fischer Jun 10 '14 at 18:02
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    This sequence arises in Analytic Number Theory. In a more general form, where we take the lcm of arithmetic progressions, I found a paper that establishes some lower bounds. The references contained talk about this sequence in particular. For instance, Hanson's 1972 paper. Here is the article I found: http://math.stanford.edu/~dankane/lcmbound.pdf – Joel Jun 10 '14 at 18:04
  • $L(n)$ is the exponent of the symmetric group $S_n$. See Landau's Function.
    Asymptotic estimates are known.     – lhf Jun 10 '14 at 18:22
  • OEIS A003418 gives many references to this sequence in the literature, some of which might address your specific question. – MJD Jun 10 '14 at 18:37

1 Answers1

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$$0\leq \Sigma \frac{l(n)}{(n!)^2}\leq\Sigma \frac{n!}{(n!)^2}=\Sigma \frac{1}{n!}=e-1 $$

Khosrotash
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