Hey,

the following function \( f_n \) iapproaches - if the limit \( n\to\infty \) exists - the intuitive logarithm to base \( b \), i.e. the intuitive Abel function of \( bx \) developed at 1:

\( f_n(x)=-\sum_{k=1}^n \left(n\\k\right)(-1)^{k}\frac{1-x^k}{1-b^k} \)

The question is whether this is indeed the logarithm, i.e. if \( \lim_{n\to\infty} f_n(x) = \log_b(x) \) for \( \left|1-\frac{x}{b}\right|<1 \), provided that the limit exists at all.

It has a certain similarity to Euler's false logarithm series (pointed out by Gottfried here) as it can indeed be proven that \( f(b^m) = m \) for natural numbers \( m \) (even for \( m=0 \) in difference to Euler's series):

\( f_n(b^m) = -\sum_{k=1}^n \left(n\\k\right)(-1)^{k}\frac{1-b^{mk}}{1-b^k} \)

if we now utilize that \( \frac{1-y^m}{1-y}=\sum_{i=0}^{m-1} y^i \) for \( y=b^k \) then we get

\(

f_n(b^m) = -\sum_{k=1}^n \left( n \\ k \right) (-1)^{k}\sum_{i=0}^{m-1} b^{ki}

= \sum_{i=0}^{m-1}\left(1-\sum_{k=0}^n \left(n\\k\right)(-1)^{k} b^{ki}\right) \)

\( f_n(b^m)=\sum_{i=0}^{m-1} 1-(1-b^i)^n \)

Hence

\( \lim_{n\to\infty} f_n(b^m) = m \)

But is this true also for non-integer \( m \)? Do we have some rules like \( \lim_{n\to\infty} f_n(x^n)=n \lim_{n\to\infty} f_n(x) \), or even \( \lim_{n\to\infty} f_n(xy)=\lim_{n\to\infty} f_n(x) + f_n(y) \)?

the following function \( f_n \) iapproaches - if the limit \( n\to\infty \) exists - the intuitive logarithm to base \( b \), i.e. the intuitive Abel function of \( bx \) developed at 1:

\( f_n(x)=-\sum_{k=1}^n \left(n\\k\right)(-1)^{k}\frac{1-x^k}{1-b^k} \)

The question is whether this is indeed the logarithm, i.e. if \( \lim_{n\to\infty} f_n(x) = \log_b(x) \) for \( \left|1-\frac{x}{b}\right|<1 \), provided that the limit exists at all.

It has a certain similarity to Euler's false logarithm series (pointed out by Gottfried here) as it can indeed be proven that \( f(b^m) = m \) for natural numbers \( m \) (even for \( m=0 \) in difference to Euler's series):

\( f_n(b^m) = -\sum_{k=1}^n \left(n\\k\right)(-1)^{k}\frac{1-b^{mk}}{1-b^k} \)

if we now utilize that \( \frac{1-y^m}{1-y}=\sum_{i=0}^{m-1} y^i \) for \( y=b^k \) then we get

\(

f_n(b^m) = -\sum_{k=1}^n \left( n \\ k \right) (-1)^{k}\sum_{i=0}^{m-1} b^{ki}

= \sum_{i=0}^{m-1}\left(1-\sum_{k=0}^n \left(n\\k\right)(-1)^{k} b^{ki}\right) \)

\( f_n(b^m)=\sum_{i=0}^{m-1} 1-(1-b^i)^n \)

Hence

\( \lim_{n\to\infty} f_n(b^m) = m \)

But is this true also for non-integer \( m \)? Do we have some rules like \( \lim_{n\to\infty} f_n(x^n)=n \lim_{n\to\infty} f_n(x) \), or even \( \lim_{n\to\infty} f_n(xy)=\lim_{n\to\infty} f_n(x) + f_n(y) \)?