Prove that: $ \sum_{r=0}^n (-1)^r (^nCr_) \frac{1 + r\log_e10}{(1 + log_e10^n)^r} = 0 $

Question

Prove that:

$$ \sum_{r=0}^n (-1)^r (^nCr_) \frac{1 + r\log_e10}{(1 + log_e10^n)^r} = 0 $$

Attempt:

$$ (1 + x)^n = C_0 + C_1x + C_2x^2 + ... + C_nx^n $$ $$ (1 + x)^n = \sum_{r=0}^nC_r(x^r)$$ $$ (1 + \frac{-1}{1 + log_e10^n})^n = \sum_{r=0}^n C_r(-1)^n\frac{1}{(1 + log_e10^n)^r}$$

I am unable to figure out how to resolve the numerator part in the question $ (1 + rlog_e10) $ into the sum. Clearly, the $ r $ here is not a power (except inside the logarithm).

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[#ffd,10px]{\ds{\sum_{r = 0}^{n}\pars{-1}^{r}\pars{^{n}Cr}\,{1 + r\log_{\expo{}}\pars{10} \over \bracks{1 + \log_{\expo{}}\pars{10^{n}}}^{r}}}} \\[5mm] = & \sum_{r = 0}^{n}{n \choose r}\bracks{-\,{1 \over 1 + \log_{\expo{}}\pars{10^{n}}}}^{r} + \left.\log_{\expo{}}\pars{10}\,\partiald{}{x} \sum_{r = 0}^{n}{n \choose r} \bracks{-\,{x \over 1 + \log_{\expo{}}\pars{10^{n}}}}^{r} \,\right\vert_{\ x\ =\ 1} \\[5mm] = &\ \bracks{1 - {1 \over 1 + \log_{\expo{}}\pars{10^{n}}}}^{n} + \left.\log_{\expo{}}\pars{10}\,\partiald{}{x}\bracks{1 - {x \over 1 + \log_{\expo{}}\pars{10^{n}}}}^{n}\,\right\vert_{\ x\ =\ 1} \\[5mm] = &\ {\log_{\expo{}}^{n}\pars{10^{n}} \over \bracks{1 + \log_{\expo{}}\pars{10^{n}}}^{n}} + \left.\log_{\expo{}}\pars{10}\,n\bracks{1 - {x \over 1 + \log_{\expo{}}\pars{10^{n}}}}^{n - 1}\bracks{-\,{1 \over 1 + \log_{\expo{}}\pars{10^{n}}}}\,\right\vert_{\ x\ =\ 1} \\[5mm] = &\ {\log_{\expo{}}^{n}\pars{10^{n}} \over \bracks{1 + \log_{\expo{}}\pars{10^{n}}}^{n}} - n\,{\log_{\expo{}}\pars{10}\log_{\expo{}}^{n - 1}\pars{10^{n}} \over \bracks{1 + \log_{\expo{}}\pars{10^{n}}}^{n}} \\[5mm] = &\ {n^{n}\log_{\expo{}}^{n}\pars{10} \over \bracks{1 + \log_{\expo{}}\pars{10^{n}}}^{n}} - n\,{\log_{\expo{}}\pars{10}\bracks{n^{n - 1}\log_{\expo{}}^{n - 1}\pars{10}} \over \bracks{1 + \log_{\expo{}}\pars{10^{n}}}^{n}} = \bbx{0} \end{align}

Answer 2

Two aspects:

• The claim is valid for positive integer $n$, but not for $n=0$ where evaluation of the left-hand side gives $1$.

• When looking at the formula we might assume that properties of the logarithm are essential, especially the rule $\log z^n=n\log z$, but neither the base nor the argument seem to be relevant for the claim. So, we can show a somewhat more general formula.

We obtain for integral $n\geq 0$ and $z>0$

\begin{align*} \color{blue}{\sum_{r=0}^n}&\color{blue}{(-1)^r\binom{n}{r}\frac{1+r\log z}{(1+\log z^n)^r}}\\ &=\sum_{r=0}^n(-1)^r\binom{n}{r}\frac{1}{\left(1+\log z^n\right)^r} +\sum_{r=1}^n(-1)^r\binom{n-1}{r-1}\frac{n\log z}{\left(1+\log z^n\right)^r}\tag{1}\\ &=1+\sum_{r=1}^{n-1}(-1)^r\left(\binom{n-1}{r}+\binom{n-1}{r-1}\right)\frac{1}{\left(1+\log z^n\right)^r}\\ &\qquad+(-1)^n\frac{1}{(1+\log z^n)^r}-\sum_{r=0}^{n-1}(-1)^r\binom{n-1}{r}\frac{\log z^n}{\left(1+\log z^n\right)^{r+1}}\tag{2}\\ &=1+\sum_{r=1}^{n-1}(-1)^r\binom{n-1}{r}\frac{1}{\left(1+\log z^n\right)^r} -\sum_{r=0}^{n-2}(-1)^r\binom{n-1}{r}\frac{1}{\left(1+\log z^n\right)^{r+1}}\\ &\qquad+(-1)^n\frac{1}{(1+\log z^n)^r}-\sum_{r=0}^{n-1}(-1)^r\binom{n-1}{r}\frac{\log z^n}{\left(1+\log z^n\right)^{r+1}}\tag{3}\\ &=1+\sum_{r=1}^{n-1}(-1)^r\binom{n-1}{r}\frac{1}{\left(1+\log z^n\right)^r}+(-1)^n\frac{1}{(1+\log z^n)^r}\\ &\qquad-\sum_{r=0}^{n-2}(-1)^r\binom{n-1}{r}\frac{1}{\left(1+\log z^n\right)^{r}}+(-1)^n\frac{\log z^n}{(1+\log z^n)^n}\tag{4}\\ &=1+(-1)^{n-1}\frac{1}{(1+\log z^n)^{n-1}}+(-1)^n\frac{1}{(1+\log z^n)^{n-1}}\\ &\qquad-1+(-1)^n\frac{\log z}{(1+\log z^n)^n}\tag{5}\\ &\,\,\color{blue}{=0}\tag{6} \end{align*} and the (generalised) claim follows.

Comment:

• In (1) we split the sum according to the terms in the numerator and use the binomial identity $q\binom{p}{q}=p\binom{p-1}{q-1}$.

• In (2) we separate the first and last term of the left sum and apply the binomial identity $\binom{p}{q}=\binom{p-1}{q}+\binom{p-1}{q-1}$. We also shift the index of the right sum to start with $r=0$.

• In (3) we split the left sum and shift the right part of it to start with $r=0$.

• In (4) we separate the last term $r=n-1$ of the right-most sum and merge the second and the right-most sum. This way we can also cancel $1+\log z^n$.

• In (5) we cancel the terms of the sums with $1\leq r\leq n-2$.

• In (6) we observe that all other terms cancel as well.