# Is this summation solvable? S_n=sum_(i=2)^n log_i(n) Should I use the derivative of log_i(n)?

Is this summation solvable? ${S}_{n}=\sum _{i=2}^{n}{\mathrm{log}}_{i}\left(n\right)$
Is it possible to solve a summation with a variable base of log?
${S}_{n}=\sum _{i=2}^{n}{\mathrm{log}}_{i}\left(n\right)$
Should I use the derivative of ${\mathrm{log}}_{i}\left(n\right)$
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Collin Gilbert
While I don't believe there is a nice closed form for ${S}_{n}$, you can write the sum in terms of known functions and constants up to a very small error. Specifically,
$\sum _{i=2}^{N}{\mathrm{log}}_{i}\left(N\right)=\text{li(N)}\mathrm{log}N+C\mathrm{log}N+O\left(1\right),$
where $\text{li}\left(N\right)$ is the logarithmic integral and $C$ is a constant equal to
$C=\frac{1}{\mathrm{log}2}+{\int }_{2}^{\mathrm{\infty }}\frac{\left\{x\right\}}{x{\mathrm{log}}^{2}x}dx.$
Proof: Writing
$\sum _{i=2}^{N}{\mathrm{log}}_{i}\left(N\right)=\mathrm{log}N\sum _{i=2}^{N}\frac{1}{\mathrm{log}i},$
our goal is then to find an asymptotic for the sum of $1/\mathrm{log}i$. Writing this as a Riemann Stieltjies integral we have
$\sum _{i=2}^{N}\frac{1}{\mathrm{log}i}={\int }_{{2}^{-}}^{{N}^{+}}\frac{1}{\mathrm{log}x}d\left[x\right]={\int }_{2}^{N}\frac{1}{\mathrm{log}x}dx-{\int }_{{2}^{-}}^{{N}^{+}}\frac{1}{\mathrm{log}x}d\left\{x\right\}.$
By integration by parts,
${\int }_{{2}^{-}}^{{N}^{+}}\frac{1}{\mathrm{log}x}d\left\{x\right\}=\frac{\left\{x\right\}}{\mathrm{log}x}{|}_{x={2}^{-}}^{x={N}^{+}}+{\int }_{2}^{N}\frac{\left\{x\right\}}{x{\mathrm{log}}^{2}x}dx$
$=\frac{1}{\mathrm{log}2}+{\int }_{2}^{\mathrm{\infty }}\frac{\left\{x\right\}}{x{\mathrm{log}}^{2}x}dx-{\int }_{N}^{\mathrm{\infty }}\frac{\left\{x\right\}}{x{\mathrm{log}}^{2}x}dx,$
and since
${\int }_{N}^{\mathrm{\infty }}\frac{\left\{x\right\}}{x{\mathrm{log}}^{2}x}dx=O\left(\frac{1}{\mathrm{log}N}\right),$
we have that
$\sum _{i=2}^{N}\frac{1}{\mathrm{log}i}=\text{li}\left(N\right)+C+O\left(\frac{1}{\mathrm{log}N}\right)$
where $\text{li}\left(N\right)$ is the logarithmic integral and
$C=\frac{1}{\mathrm{log}2}+{\int }_{2}^{\mathrm{\infty }}\frac{\left\{x\right\}}{x{\mathrm{log}}^{2}x}dx.$
Thus it follows that
$\sum _{i=2}^{N}{\mathrm{log}}_{i}\left(N\right)=\text{li(N)}\mathrm{log}N+C\mathrm{log}N+O\left(1\right).$
(Note that the asymptotic is then $\sum _{i=2}^{N}{\mathrm{log}}_{i}\left(N\right)\sim N.$)
Remark: In fact, we could apply integration by parts again to work out the $O\left(1\right)$ term exactly and evaluate the sum up to an error of $O\left(\frac{1}{N}\right)$. This general process of writing the sum $\sum _{k\le N}f\left(k\right)$ as a series whose main term is ${\int }_{1}^{N}f\left(x\right)dx$ is known as Euler-Maclaurin summation.
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deksteenk9
Since ${\mathrm{log}}_{i}n=\frac{\mathrm{log}n}{\mathrm{log}i}$, we have
${S}_{n}=\sum _{i=2}^{n}\frac{\mathrm{log}n}{\mathrm{log}i}=\mathrm{log}n\sum _{i=2}^{n}\frac{1}{\mathrm{log}i}$
and we have by Euler-McLaurin summation formula
$\sum _{i=2}^{n}\frac{1}{\mathrm{log}i}={\int }_{2}^{n}\frac{\text{d}x}{\mathrm{log}x}+\mathrm{log}\sqrt{2n}+O\left(\frac{1}{\mathrm{log}n}\right),$
${S}_{n}=\text{Li}\left(n\right)\mathrm{log}n+\frac{1}{2}{\mathrm{log}}^{2}n+\mathrm{log}\sqrt{2}\mathrm{log}n+O\left(1\right).$
Hope this helps.
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