Jay Mckay

2022-01-24

How can I calculate the following sum involving binomial terms:
$\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{\left(-1\right)}^{k}}{{\left(k+1\right)}^{2}}$

### Answer & Explanation

lirwerwammete9t

We have
$\frac{1}{z}{\int }_{0}^{z}\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){s}^{k}ds=\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{z}^{k}}{k+1}$
so that
$-{\int }_{0}^{z}\frac{1}{t}{\int }_{0}^{t}\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){s}^{k}dsdt=-\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{z}^{k+1}}{{\left(k+1\right)}^{2}}$
Setting $z=-1$ gives an expression for your sum,
$\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{\left(-1\right)}^{k}}{{\left(k+1\right)}^{2}}={\int }_{-1}^{0}\frac{1}{t}{\int }_{0}^{t}\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){s}^{k}dsdt$
Now, $\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){s}^{k}={\left(1+s\right)}^{n}$, so
$\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{\left(-1\right)}^{k}}{{\left(k+1\right)}^{2}}={\int }_{-1}^{0}\frac{1}{t}{\int }_{0}^{t}{\left(1+s\right)}^{n}dsdt$
$=\frac{1}{n+1}{\int }_{-1}^{0}\frac{1}{t}\left[{\left(1+t\right)}^{n+1}-1\right]dt$
$=\frac{1}{n+1}{\int }_{0}^{1}\frac{{u}^{n+1}-1}{u-1}du$
$=\frac{1}{n+1}{\int }_{0}^{1}\sum _{k=0}^{n}{u}^{k}du$

Brenton Pennington

One application of the absorption identity gets one of the factors of $k+1$ out of the denominator:
$\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{\left(-1\right)}^{k}}{{\left(k+1\right)}^{2}}=\frac{1}{n+1}\sum _{k=0}^{n}\left(\begin{array}{c}n+1\\ k+1\end{array}\right)\frac{{\left(-1\right)}^{k}}{k+1}$
$=\frac{1}{n+1}\sum _{k=1}^{n+1}\left(\begin{array}{c}n+1\\ k\end{array}\right)\frac{{\left(-1\right)}^{k+1}}{k}$
By using the basic binomial coefficient recursion formula, we can make that happen.
Let $f\left(n\right)=\sum _{k=1}^{n}\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{\left(-1\right)}^{k+1}}{k}$ Then looking at the difference of $f\left(n+1\right)$ and $f\left(n\right)$ gives us
$f\left(n+1\right)-f\left(n\right)=\sum _{k=1}^{n+1}\left(\begin{array}{c}n+1\\ k\end{array}\right)\frac{{\left(-1\right)}^{k+1}}{k}-\sum _{k=1}^{n}\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{\left(-1\right)}^{k+1}}{k}$
$=\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{\left(-1\right)}^{k}}{k+1}$
$=\frac{1}{n+1}\sum _{k=0}^{n}\left(\begin{array}{c}n+1\\ k+1\end{array}\right){\left(-1\right)}^{k}$
$=\frac{1}{n+1}\left(1+\sum _{k=0}^{n+1}\left(\begin{array}{c}n+1\\ k\end{array}\right){\left(-1\right)}^{k+1}\right)$
$=\frac{1}{n+1}$
where in the last step we used the fact that the alternating sum of the binomial coefficients is 0.
Thus
$f\left(n+1\right)=\sum _{k=0}^{n}\left(f\left(k+1\right)-f\left(k\right)\right)=\sum _{k=0}^{n}\frac{1}{k+1}={H}_{n+1}$
Therefore,

RizerMix

$\sum _{k=0}^{n}\left(\left(n\right),\left(k\right)\right)\frac{\left(-1{\right)}^{k}}{\left(k+1{\right)}^{2}}$ $=-\sum _{k=0}^{n}\left(\left(n\right),\left(k\right)\right)\left(-1{\right)}^{k}{\int }_{0}^{1}\mathrm{ln}\left(x\right){x}^{k}dx$ $=-{\int }_{0}^{1}\mathrm{ln}\left(x\right)\sum _{k=0}^{n}\left(\left(n\right),\left(k\right)\right)\left(-x{\right)}^{k}dx$ $=-{\int }_{0}^{1}\mathrm{ln}\left(x\right)\left(1-x{\right)}^{n}dx$ $=-\frac{d}{d\mu }{\int }_{0}^{1}{x}^{\mu }\left(1-x{\right)}^{n}dx{|}_{\mu =0}$ $=-\frac{d}{d\mu }\frac{\mathrm{\Gamma }\left(\mu +1\right)\mathrm{\Gamma }\left(n+1\right)}{\mathrm{\Gamma }\left(\mu +n+2\right)}{|}_{\mu =0}=\frac{{H}_{n+1}}{n+1}$ Note that $\mathrm{\Gamma }\left(a+\mu \right)=\mathrm{\Gamma }\left(a\right)\left[1+\left({H}_{a-1}-\gamma \right)\mu +O\left({\mu }^{2}\right)\right]$ such that $=-\frac{\mathrm{\Gamma }\left(\mu +1\right)\mathrm{\Gamma }\left(n+1\right)}{\mathrm{\Gamma }\left(\mu +n+2\right)}=-\frac{n!}{\left(n+1\right)!}×\left[1+\left[\left(1-\gamma \right)-\left({H}_{n+1}-\gamma \right)\right]\mu +O\left({\mu }^{2}\right)\right]$ $=-\frac{1}{n+1}+\frac{{H}_{n}}{n+1}\mu +O\left({\mu }^{2}\right)$ with ${H}_{0}=0$

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