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Presumably \begin{align} \operatorname{Li}(x) = & \sum_{n = 2}^{x} \dfrac{1}{\log(n)}+ O(\log(x))\\ \end{align} where \begin{align} \operatorname{Li}(x) = & \int_{2}^{x}\dfrac{1}{\log(t)}\operatorname{d}t \end{align} By Euler-Maclaurin approximation \begin{align} \sum_{n = 2}^{x} \dfrac{1}{\log(n)} \approx & \int_{2}^{x} \dfrac{1}{\log(t)} \operatorname{d}t + \dfrac{1}{2}\left(\dfrac{1}{\log(2)} + \dfrac{1}{\log(x)}\right) \\ \end{align} and \begin{align} &\dfrac{1}{2}\left(\dfrac{1}{\log(2)} + \dfrac{1}{\log(x)}\right)=O(\log(x)) \end{align}

I am having greater difficulty calculating the error term ($e$) for

\begin{align} \operatorname{Li}(x)=\sum_{n=1}^{x}\dfrac{1}{\sum_{k=1}^{n}k^{-1}}+e \end{align}

though, since it is a nested sum. I looked here, but am struggling a little.

Community
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martin
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1 Answers1

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The difference: $$\int_{2}^{x}\frac{dt}{\log t}-\sum_{n=2}^{x}\frac{1}{\log n},$$ regarded as the approximation error for a Riemann sum, is bounded by: $$\sum_{n=2}^{x}\left|\frac{1}{\log n}-\frac{1}{\log(n+1)}\right|\leq\sum_{n=2}^{x}\frac{1}{n\log^2 n}\leq\frac{1}{\log^2 2}\sum_{n=2}^{x}\frac{1}{n}=O(\log x).$$ Moreover: $$H_n = \sum_{k=1}^{n}\frac{1}{k} = \sum_{k=1}^{n}\left(\log\left(1+\frac{1}{k}\right)+O(1/k^2)\right)=\log(n+1)+O(1)=\log(n)+O(1)$$ gives that: $$\frac{1}{\log^2 n}\ll\left|\frac{1}{\log n}-\frac{1}{H_n}\right|\ll\frac{1}{\log^2 n},$$ hence you just have to estimate $$\sum_{n=2}^{x}\frac{1}{\log^2 n}$$ to find the error term in last formula.

Jack D'Aurizio
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