Integrating to derive relation between binomial co-efficient

Question

If $C_o,C_1...$ are the binomial coefficents in the expansion of $(1+x)^n$ and $$\sum_{r=0}^{n} (-1)^r \binom{n}{r} \frac{1}{(r+1)^2} = k \sum_{r=0}^{n} \frac{1}{r+1}$$ Find a k such that the above equation is satisfied

My attempt: $$ (1+x)^n = \sum_{k=0}^{k=n} \binom{n}{k} x^k$$

integrate both sides

$$ \frac{ (1+x)^{n+1} }{n+1} =\sum_{k=0}^{n} \binom{n}{k} \frac{x^{k+1}}{k+1} +C$$

$$ x= 0 $$

$$ \implies \frac{-n}{n+1} = C$$

Divide both sides by 'x' and integrate

$$ \int \frac{ (1+x)^{n+1} }{ (x) n+1} dx = \sum_{k=0}^{n} \binom{n}{k} \frac{x^{k+1}}{(k+1)^2} - \frac{n}{n+1} \ln(x)+ C'$$

Or,

$$ \int \frac{ (1+x)^{n+1} }{ (x) n+1} dx - \frac{n}{n+1} \ln(x)+ C' = \sum_{k=0}^{n} \binom{n}{k} \frac{x^{k+1}}{(k+1)^2}$$

Not so sure what do here, like what to put as bounds. Slightly concerned I may have to evaluate a negative 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[5px,#ffd]{\sum_{r = 0}^{n}\pars{-1}^{r}{n \choose r} {1 \over \pars{r + 1}^{2}}} = \sum_{r = 0}^{n}\pars{-1}^{r}{n \choose r}\ \overbrace{\bracks{-\int_{0}^{1}\ln\pars{x}\, x^{r}\,\dd x}} ^{\ds{1 \over \pars{r + 1}^{2}}} \\[5mm] = &\ -\int_{0}^{1}\ln\pars{x}\sum_{r = 0}^{n}{n \choose r}\pars{-x}^{r}\,\dd x = -\int_{0}^{1}\ln\pars{x}\pars{1 - x}^{n}\,\dd x \\[5mm] = &\ -\bracks{\xi^{1}}\int_{0}^{1}x^{\xi}\pars{1 - x}^{n}\,\dd x = -\bracks{\xi^{1}}\bracks{\Gamma\pars{\xi + 1}\Gamma\pars{n + 1} \over \Gamma\pars{\xi + n + 2}} \\[5mm] = &\ -n!\bracks{\xi^{1}}\bracks{% \Gamma\pars{1} + \Gamma\, '\pars{1}\xi \over \Gamma\pars{n + 2} + \Gamma\, '\pars{n + 2}\xi} \\[5mm] = &\ -\,{1 \over n + 1}\bracks{\xi^{1}}\bracks{% 1 - \gamma\xi \over 1 + \Psi\pars{n + 2}\xi} \\[5mm] = &\ -\,{1 \over n + 1}\bracks{\xi^{1}}\braces{\vphantom{\Large A}% \pars{\vphantom{\large A}1 - \gamma\xi}\bracks{\vphantom{\large A}1 - \Psi\pars{n + 2}\xi}} \\[5mm] = &\ -\,{1 \over n + 1}\bracks{-\gamma - \Psi\pars{n + 2}} = -\,{1 \over n + 1}\pars{-H_{n + 1}} = \\[5mm] = &\ {1 \over n + 1}\sum_{r = 0}^{n}{1 \over r + 1} \implies \bbx{k = {1 \over n + 1}} \\ & \end{align}

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$\ds{\bracks{\xi^{m}}}$ is the Coefficient Extraction Operator.

$\ds{\Gamma}$ is the Gamma Function.

$\ds{\gamma}$ is the Euler-Mascheroni Constant.

$\ds{\Psi}$ is the Digamma Function.

$\ds{H_{z}}$ is a Harmonic Number.

Answer 2

Here is my approach. Let denote that $H_n = \sum_{k=1}^{n}\frac{1}{k}$ is the n-th Harmonic series. Then it is easy to observe that: $$H_n = \sum_{k=1}^{n}\frac{1}{k} = \sum_{k=1}^{n}\int_{0}^{1} x^{k-1}dx = \int_{0}^{1}\sum_{k=1}^{n} x^{k-1}dx = \int_{0}^{1} \frac{1-x^n}{1-x}~dx$$ For your approach above, instead of using $(1+x)^n$, why don't you use $(1-x)^n$? $$(1-x)^n = \sum_{k=0}^{n} \binom{n}{k}(-1)^r x^r \Rightarrow \int_{0}^{t} (1-x)^n ~dx = \int_{0}^{t}\sum_{k=0}^{n} \binom{n}{k}(-1)^r x^r dx$$$$\Rightarrow \frac{1- (1-t)^{n+1}}{t(n+1)}= \sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^r t^{r}}{r+1}$$ Now, notice that: $$LHS=\sum_{r=0}^{n} \binom{n}{r}\frac{(-1)^r}{(r+1)^2}=\sum_{r=0}^{n} \binom{n}{r}\frac{(-1)^rt^{r+1}}{(r+1)^2}\Bigg|_0^1=\int_{0}^{1}\sum_{r=0}^{n} \binom{n}{r}\frac{(-1)^r t^{r}}{r+1}~dt$$$$=\int_{0}^{1} \frac{1- (1-t)^{n+1}}{t(n+1)} ~ dt$$ Here we will move to the RHS: $$RHS=k \sum_{r=0}^{n} \frac{1}{r+1} = k\cdot H_{n+1} = k \cdot \int_{0}^{1} \frac{1-t^{n+1}}{1-t}~dt= k \int_{0}^{1} \frac{1-(1-t)^{n+1}}{t}~dt \cdot $$ By comparing $LHS$ with $RHS$, we obtain that $k=\frac{1}{n+1}$

Answer 3

In trying to evaluate

$$S_n = \sum_{r=0}^n (-1)^r {n\choose r} \frac{1}{(r+1)^2}$$

we introduce

$$f(z) = \frac{(-1)^n \times n!}{(z+1)^2} \prod_{q=0}^n \frac{1}{z-q}$$

which has the property that for $0\le r\le n$

$$\mathrm{Res}_{z=r} f(z) = \frac{(-1)^n \times n!}{(r+1)^2} \prod_{q=0}^{r-1} \frac{1}{r-q} \prod_{q=r+1}^n \frac{1}{r-q} \\ = \frac{(-1)^n \times n!}{(r+1)^2} \frac{1}{r!} \frac{(-1)^{n-r}}{(n-r)!} = (-1)^r {n\choose r} \frac{1}{(r+1)^2}.$$

It follows that

$$S_n = \sum_{r=0}^n \mathrm{Res}_{z=r} f(z).$$

Now residues sum to zero and the residue at infinity of $f(z)$ is zero by inspection, therefore

$$S_n = - \mathrm{Res}_{z=-1} f(z) = (-1)^{n+1} \times n! \times \left. \left( \prod_{q=0}^n \frac{1}{z-q} \right)' \right|_{z=-1} \\ = (-1)^{n+1} \times n! \times \left. \prod_{q=0}^n \frac{1}{z-q} \sum_{q=0}^n \frac{1}{q-z} \right|_{z=-1} \\ = (-1)^{n+1} \times n! \times (-1)^{n+1} \prod_{q=0}^n \frac{1}{q+1} \times \sum_{q=0}^n \frac{1}{q+1} \\ = n! \times \frac{1}{(n+1)!} \times \sum_{q=0}^n \frac{1}{q+1} = \frac{1}{n+1} \sum_{q=0}^n \frac{1}{q+1} = \frac{1}{n+1} H_{n+1}.$$

We see that the desired factor on the sum is $k=\frac{1}{n+1}.$