Savanah Boone

Answered

2022-07-01

Simplifying this logarithm series

$\sum _{i\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}2}^{99}\frac{1}{{\mathrm{log}}_{i}(99!)}$

How would you evaluate (or at least simplify) this logarithm series?

$\sum _{i\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}2}^{99}\frac{1}{{\mathrm{log}}_{i}(99!)}$

How would you evaluate (or at least simplify) this logarithm series?

Answer & Explanation

isscacabby17

Expert

2022-07-02Added 13 answers

Note that ${\mathrm{log}}_{a}b=\frac{\mathrm{log}b}{\mathrm{log}a}$, for any choice of base of the logarithm. (Mathematicians tend to mean the natural logarithm when they write $\mathrm{log}$.)

In particular, $\frac{1}{{\mathrm{log}}_{a}b}=\frac{\mathrm{log}a}{\mathrm{log}b}={\mathrm{log}}_{b}a$.

Therefore your sum is:

$\sum _{i=2}^{99}{\mathrm{log}}_{99!}i={\mathrm{log}}_{99!}\left(\prod _{i=2}^{99}i\right)={\mathrm{log}}_{99!}99!=1$

where in first equality I used the well-known fact that $\sum \mathrm{log}=\mathrm{log}\prod $.

In particular, $\frac{1}{{\mathrm{log}}_{a}b}=\frac{\mathrm{log}a}{\mathrm{log}b}={\mathrm{log}}_{b}a$.

Therefore your sum is:

$\sum _{i=2}^{99}{\mathrm{log}}_{99!}i={\mathrm{log}}_{99!}\left(\prod _{i=2}^{99}i\right)={\mathrm{log}}_{99!}99!=1$

where in first equality I used the well-known fact that $\sum \mathrm{log}=\mathrm{log}\prod $.

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