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Is there any closed formula for the following summation?

$$\sum_{k=2}^n \frac{1}{\log_2(k)}$$

Git Gud
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1 Answers1

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There is no closed form as such. However, you can use the Abel summation technique from here to derive the asymptotic. We have \begin{align} S_n & = \sum_{k=2}^n \dfrac1{\log(k)} = \int_{2^-}^{n^+} \dfrac{d \lfloor t \rfloor}{\log(t)} = \dfrac{n}{\log(n)} - \dfrac2{\log(2)} + \int_2^{n} \dfrac{dt}{\log^2(t)}\\ & =\dfrac{n}{\log(n)} - \dfrac2{\log(2)} + \int_2^{3} \dfrac{dt}{\log^2(t)} + \int_3^{n} \dfrac{dt}{\log^2(t)}\\ &\leq \dfrac{n}{\log(n)} \overbrace{- \dfrac2{\log(2)} + \int_2^{3} \dfrac{dt}{\log^2(t)}}^{\text{constant}} + \dfrac1{\log(3)}\int_3^{n} \underbrace{\dfrac{dt}{\log(t)}}_{\leq S_n}\\ \end{align} We have get that $$S_n \leq \dfrac{n}{\log(n)} + \text{constant} + \dfrac{S_n}{\log(3)} \implies S_n \leq \dfrac{\log(3)}{\log(3)-1} \left(\dfrac{n}{\log(n)} + \text{constant}\right) \tag{$\star$}$$ With a bit more effort you can show that $$S_n \sim \dfrac{n}{\log n}$$

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