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Warning: This post contains more than one question and is pretty long. I decided include them all in this post, because they all emerged from the same question. Furthermore, I decided to display all the details of my calculations, in order to explain my thoughts. They are not necessarily relevant; you might skip some parts.

I'm currently trying to evaluate the following unusual sum: $$ \Xi=\sum_{b=2}^{\infty}{\left[\sum_{k=1}^{\infty}{\left(\frac{Q_b(k)}{k(k+1)}+\left(1+\frac1b\right)\frac{(-1)^k}{kb^k}\right)}\right]} $$ where $Q_b(k)$ is the digit sum of $k$ in base $b$. One part can be evaluated using the natural logarithm: $$ \Xi=\sum_{b=2}^{\infty}{\left[\left(\sum_{k=1}^{\infty}{\frac{Q_b(k)}{k(k+1)}}\right)-\left(1+\frac1b\right)\ln(b)\right]} $$ Now, define $S_b=\sum_{k=1}^{\infty}{\frac{Q_b(k)}{k(k+1)}}$ and, in order to evaluate it, $S_b(N)=\sum_{k=1}^{N}{\frac{Q_b(k)}{k(k+1)}}$. We have: $$ S_b(N)=\sum_{k=1}^{N}{\frac{Q_b(k)}{k(k+1)}}=\sum_{k=1}^{N}{Q_b(k)\left(\frac1k-\frac1{k+1}\right)}=\sum_{k=1}^{N}{\frac{Q_b(k)}{k}}-\sum_{k=1}^{N}{\frac{Q_b(k)}{k+1}}=1-\frac{Q_b(N)}{N+1}+\sum_{k=2}^{N}{\frac{Q_b(k)-Q_b(k-1)}{k}}=-\frac{Q_b(N)}{N+1}+\sum_{k=1}^{N}{\frac{Q_b(k)-Q_b(k-1)}{k}} $$ Now, we take a look at $Q_b(k)-Q_b(k-1)$.

We define the function $\delta_b:\mathbb{N_0}\to\mathbb{N_0}$ through $\delta_b(n)=r$ if $r$ is the (unique) number, such that $n\equiv-1\pmod{b^{r}}$ but $n\not\equiv-1\pmod{b^{r+1}}$. Note that $\delta_b(n)$ is the number of $(b-1)$s in the end of the base $b$ representation of $n$. Now, we define $R_b:\mathbb{N_0}\to\mathbb{N_0}$ as $$R_b(n)=\frac{n-\left(b^{\delta_b(n)}-1\right)}{b^{\delta_b(n)}}$$ such that $R_b(n)$ results by removing the $(b-1)$s at the end of $n$. So i.e. we have $\delta_{10}(21169299)=2$ and $R_{10}(21169299)=211692$.

Clearly, we have $$Q_b(n)=Q_b(R_b(n))+Q_b\left(b^{\delta_b(n)}-1\right)=Q_b(R_b(n))+(b-1)\delta_b(n)$$ and: $$ Q_b(n+1)=Q_b(R_b(n))+1 $$ since $R_b(n)$ ends not with $(b-1)$ and the $(b-1)$ tail of $n$ transforms into a string of zeros. Therefore: $$ Q_b(k)-Q_b(k-1)=1-(b-1)\delta_b(k-1) $$ Now, we can calculate $S_{b}\left(b^N-1\right)$. For $1≤k≤b^N-1$ we have $0≤\delta_b(k-1)≤N-1$. $\delta_b(k-1)=r$ is equivalent to $k\equiv0\pmod{b^{r}}$ and $k\not\equiv0\pmod{b^{r+1}}$, so $k=b^r(bn+\beta)$ for some $n≥0$ and $1≤\beta≤b-1$. We calculate $S_{b}\left(b^N-1\right)$ by summing up the terms for which $\delta_b(k-1)=0$, then the terms for which $\delta_b(k-1)=1$ and so on. This yields: $$ \begin{align} S_{b}\left(b^N-1\right) & =-\frac{Q_b(b^N-1)}{b^N}+\sum_{n=0}^{b^{N-1}-1}{\sum_{\beta=1}^{b-1}{\frac{1-(b-1)\cdot 0}{bn+\beta}}}+\sum_{n=0}^{b^{N-2}-1}{\sum_{\beta=1}^{b-1}{\frac{1-(b-1)\cdot 1}{b(bn+\beta)}}}+\dots+\sum_{n=0}^{b^{0}-1}{\sum_{\beta=1}^{b-1}{\frac{1-(b-1)\cdot (N-1)}{b^{n-1}(bn+\beta)}}} \\ & =-\frac{N(b-1)}{b^N}+\sum_{r=0}^{N-1}{\sum_{n=0}^{b^{N-r-1}-1}{\sum_{\beta=1}^{b-1}{\frac{1-(b-1)\cdot r}{b^r(bn+\beta)}}}} \\ &=-\frac{N(b-1)}{b^N}+\sum_{r=0}^{N-1}{\left(\frac{1-(b-1)\cdot r}{b^r}\right)\sum_{n=0}^{b^{N-r-1}-1}{\sum_{\beta=1}^{b-1}{\frac{1}{bn+\beta}}}} \\ &=-\frac{N(b-1)}{b^N}+\sum_{r=0}^{N-1}{\left(\frac{1-(b-1)\cdot r}{b^r}\right)\sum_{n=0}^{b^{N-r-1}-1}{\left(H_{b(n+1)}-H_{bn}-\frac{1}{b(n+1)}\right)}} \\ &=-\frac{N(b-1)}{b^N}+\sum_{r=0}^{N-1}{\left(\frac{1-(b-1)\cdot r}{b^r}\right)\left(H_{b^{N-r}}-\frac{1}{b}H_{b^{N-r-1}}\right)}\\ &=-\frac{N(b-1)}{b^N}+\sum_{r=0}^{N-1}{\left(\frac{1-(b-1)\cdot r}{b^r}\right)H_{b^{N-r}}}-\sum_{r=0}^{N-1}{\left(\frac{1-(b-1)\cdot r}{b^{r+1}}\right)H_{b^{N-r-1}}}\\ &=-\frac{N(b-1)}{b^N}+\sum_{r=0}^{N-1}{\left(\frac{1-(b-1)\cdot r}{b^r}\right)H_{b^{N-r}}}-\sum_{r=1}^{N}{\left(\frac{1-(b-1)\cdot (r-1)}{b^{r}}\right)H_{b^{N-r}}}\\ &=-\frac{N(b-1)}{b^N}+\sum_{r=0}^{N-1}{\left(\frac{1-(b-1)\cdot r}{b^r}\right)H_{b^{N-r}}}-\sum_{r=1}^{N}{\left(\frac{1-(b-1)\cdot r}{b^{r}}\right)H_{b^{N-r}}}-\sum_{r=1}^{N}{\left(\frac{b-1}{b^{r}}\right)H_{b^{N-r}}}\\ &=-\frac{N(b-1)}{b^N}+H_{b^{N}}-\frac{1-(b-1)\cdot N}{b^{N}} -\sum_{r=1}^{N}{\left(\frac{b-1}{b^{r}}\right)H_{b^{N-r}}}\\ &=-\frac{1}{b^N}+H_{b^{N}}-\sum_{r=1}^{N}{\left(\frac{b-1}{b^{r}}\right)H_{b^{N-r}}}\\ &=-\frac{1}{b^N}+H_{b^{N}}-\sum_{r=0}^{N-1}{\left(\frac{b-1}{b^{N-r}}\right)H_{b^{r}}}\\ &=-\frac{1}{b^N}+H_{b^{N}}-\frac{b-1}{b^{N}}\sum_{r=0}^{N-1}{b^rH_{b^{r}}}\\ \end{align} $$ That $S_b$ converges is easy to see, so $S_b=\lim_{N\to\infty}{S_b\left(b^N-1\right)}$. But so far, I was unable to compute the limit. Using Wolfram alpha though, I conjectured that $S_b=\frac{b}{b-1}\ln(b)$, which seems to hold numerically. Using this conjecture, we could modify the original sum as follows: $$ \Xi=\sum_{b=2}^{\infty}{\left[\frac{b}{b-1}\ln(b) -\left(1+\frac1b\right)\ln(b)\right]}=\sum_{b=2}^{\infty}{\frac{\ln(b)}{b(b-1)}}=\sum_{b=1}^{\infty}{\frac{\ln(b+1)}{b(b+1)}}=\sum_{b=1}^{\infty}{\ln(b+1)\left(\frac{1}{b}-\frac{1}{b+1}\right)}=\sum_{b=1}^{\infty}{\frac{1}{b}\ln\left(1+\frac{1}{b}\right)}=\sum_{b=1}^{\infty}{\sum_{k=1}^{\infty}{\frac{(-1)^{k+1}}{kb^{k+1}}}}=\sum_{k=1}^{\infty}{\frac{(-1)^{k+1}}{k}\zeta(k+1)} $$ Or on the other hand: $$ \Xi=\sum_{b=2}^{\infty}{\frac{\ln(b)}{b(b-1)}}=\sum_{b=2}^{\infty}{\sum_{k=2}^{\infty}{\frac{\ln(b)}{b^k}}}=\sum_{k=2}^{\infty}-\zeta'(k) $$ Is it possible to calculate one of these sums?

So to sum up: How to calculate the limit $S_b=\lim_{N\to\infty}{S_b\left(b^N-1\right)}$? Is it true that $S_b=\frac{b}{b-1}\ln(b)$? Can we find a closed form for $\Xi$? Thanks in advance!

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