How to evaluate the sum $\displaystyle\sum \limits_{n=1}^\infty \frac1{L_n}$? Where $L_n$ is the least common multiple of $1, 2, 3,\ldots, n$, e.g. least common multiple of $(6,3,4)$ is $12$.
Evaluate $\sum \limits_{n=1}^\infty \frac1{L_n}$ where $L_n$ is least common multiple of $1, 2, 3,\ldots,n$
11
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real-analysis
sequences-and-series
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4I'm not convinced there is a closed form; http://oeis.org/A064859. Convergence follows from $LCM(1,2,\dots,n) \geq n(n-1)$ and $\sum_{n=2}^{\infty} \frac{1}{n(n-1)} < \infty$ by easy telescoping – 2012-02-05
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4I agree about closed form...the Prime Number Theorem is equivalent to the fact that $\psi(n) \sim n,$ where $\psi(n)= \log \mbox{lcm} (1,2,3,\ldots,n)$ is the Chebyshev function of the second kind, see equation (22.1.3) on page 340 of Hardy and Wright and https://oeis.org/A003418 Anyway, your denominator is, very roughly, $e^n,$ so convergence of the sum is quick, as in a geometric series. Also Theorem 420 on page 345 and Theorem 434 on page 362 of Hardy and Wright. – 2012-02-05
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4Summing the series to $n=120$ and then to $n=130$ gives the same result to $20$ places, namely $1.7877804561724665461$. This number is not listed in Plouffe's inverter. – 2012-02-05
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0...neither is the reciprocal, $0.5593527977931594274755\dots$ – 2012-02-05
1 Answers
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Convergence follows from $LCM(1,2,…,n)≥n(n−1)$ and $\sum_{n=2}^{\infty} \frac{1}{n(n-1)} < \infty$