Evaluating value of $$\sum^{19}_{k=1}\frac{(-1)^{k-1}k}{\binom{20}{k}}$$
I am trying to solve it using Integration.
$$\int^1_0 t^{k}(1-t)^{n-k}dt=\frac{1}{n+1}\cdot \frac{1}{\binom{n}{k}}$$
So our sum is
$\displaystyle =21\sum^{19}_{k=1}(-1)^{k-1}k\int^1_0(t)^{k}(1-t)^{n-k}dt$
$\displaystyle =21\int^1 \sum^{19}_{k=1}(-1)^{k-1}t^{k}{(1-t)^{n-k}}dt$
$\displaystyle =-21\int^1_0\sum^{19}_{k=1}(-1)^{k}k\bigg(\frac{t}{1-t}\bigg)^{k}dt$
I did not know how do i solve it, Help me
Define: $$f( x) =\sum _{k=0}^{n}\frac{x^{k-1} \cdot k}{\binom{n}{k}}$$
The sum you want to evaluate is $f(-1)+n(-1)^n$ for $n=20$.
By formula of finite geometric series: $$\sum_{k=0}^nx^k=\frac{x^{n+1}-1}{x-1}$$
Differentiating this once w.r.t. $x$, $$\sum_{k=0}^nx^{k-1}\cdot k=\underbrace{\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}}_{:=\ g(x)}$$
Define the R.H.S. as $g(x)$. Now substitute $x=-\frac{1-t}{t}$ and multiply both sides by $(n+1)(1-t)t^{n-1}$. $$(n+1)\sum_{k=0}^nk(-1)^{k-1}\cdot t^{n-k}\cdot (1-t)^{k}\\=(n+1)(1-t)t^{n-1}\cdot g\left(-\frac{1-t}{t}\right)$$
Now integrate both sides w.r.t. $t$ from $0$ to $1$. Using the fact that $\displaystyle\int^1_0 t^{k}(1-t)^{n-k}\mathrm dt=\frac{1}{n+1}\cdot \frac{1}{\binom{n}{k}},\tag*{}$ the LHS becomes $f(-1)$ and the RHS is a seemingly complicated integral (which is elementary, though tedious). I found that:
$$f(-1)=-\frac{(n+1)((n^2+3n+1)(-1)^n-1)}{(n+2)(n+3)}$$
I double checked this answer for different values of $n$ using Python. It makes sense numerically.
Overall, it was tedious exercise which seems to be of little use.
Here we continue OP's approach using a formula for reciprocal binomial coefficients \begin{align*} \binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}dz\tag{1} \end{align*} followed by geometric series expansion, taking derivatives and applying partial integration.
We obtain \begin{align*} \color{blue}{\sum_{k=1}^{19}}&\color{blue}{(-1)^{k-1}k\binom{20}{k}^{-1}}\\ &=21\sum_{k=1}^{19}(-1)^{k-1}k\int_0^1z^k(1-z)^{20-k}\,dz\tag{2}\\ &=-21\int_0^1(1-z)^{20}\sum_{k=1}^{19}(-1)^{k-1}k\left(\frac{z}{1-z}\right)^k\,dz\tag{3}\\ &=-21\int_0^1(1-z)^{20}\sum_{k=0}^{19}(-1)^kz(1-z)\left.\left(\frac{d}{dt}\left(\frac{t}{1-t}\right)^k\right)\right|_{t=z}\,dz\tag{4}\\ &=-21\int_0^1z(1-z)^{21}\left.\left(\frac{d}{dt}\sum_{k=0}^{19}\left(-\frac{t}{1-t}\right)^k\right)\right|_{t=z}\,dz\\ &=-21\int_0^1z(1-z)^{21}\left.\left(\frac{d}{dt}\left(\frac{1-\left(-\frac{t}{1-t}\right)^{20}}{1-\left(-\frac{t}{1-t}\right)}\right)\right)\right|_{t=z}\,dz\tag{5}\\ &=21\int_{0}^1z(1-z)^{21}\left.\left(1-\frac{t^{19}(t-20)}{(t-1)^{20}}\right)\right|_{t=z}\,dz\tag{6}\\ &=21\int_{0}^1z(1-z)^{21}\left(1-\frac{z^{19}(z-20)}{(z-1)^{20}}\right)\,dz\\ &=21\int_{0}^1z(1-z)^{21}\,dz-21\int_{0}^{1}z^{20}(1-z)(z-20)\,dz\tag{7}\\ &=21\left(-z\left.\frac{1}{22}(1-z)^{22}\right|_{0}^1+\frac{1}{22}\int_{0}^1(1-z)^{22}\,dz\right)\\ &\qquad-21\int_0^1\left(-20z^{20}+21z^{21}-z^{22}\right)\,dz\\ &=\left.-\frac{21}{22\cdot23}(1-z)^{23}\right|_{0}^1 -21\left(-\frac{20}{21}+\frac{21}{22}-\frac{1}{23}\right)\\ &=\frac{21}{22\cdot 23}+20-\frac{21\cdot 21}{22}+\frac{21}{23}\\ &\,\,\color{blue}{=\frac{10}{11}} \end{align*}
Comment:
In (2) we write the reciprocal of a binomial coefficient according to (1) which is based upon the beta function.
In (3) we factor out terms from the sum independent of $k$.
In (4) we get rid of the factor $k$ by using the derivative operator. Here we use the relation \begin{align*} z(1-z)\frac{d}{dz}\left(\frac{z}{1-z}\right)^k=k\left(\frac{z}{1-z}\right)^k \end{align*}
In (5) we use the finite geometric sum formula and do some simplifications in the next step.
In (6) we simplify and differentiate with respect to $t$.
In (7) we apply partial integration to the left-hand integral.
Let $$S = \sum^{19}_{k=1}\frac{(-1)^{k-1}k}{\binom{20}{k}}$$
Now,
$S = \frac{1}{\binom {20}{1}} - \frac{2}{\binom {20}{2}}.....+\frac{19}{\binom {20}{19}}$ ......$(1)$ $S = \frac{19}{\binom {20}{1}} - \frac{18}{\binom {20}{2}}.....+\frac{1}{\binom {20}{19}}$ ......$(2)$
Note that equation ($2$) is obtained by writing the original series in reverse order and using the fact $\binom{n}{k} = \binom{n}{n-k}$.
Now , adding $(1)$ and $(2)$ , we have :
$$ S = 10S'- \binom{20}{10}$$
Where,
$$S' = \sum_{k=1}^9 \frac{(-1)^{k-1}}{\binom {20}{k}}$$
For easily evaluating $S'$ , you can refer to this post.
Let us calculate the sum more generally
$$s(n)=\sum _{k=1}^n \frac{(-1)^{k-1} k}{\binom{n+1}{k}}, n=1,2,3,...\tag{1}$$
The result can be written as
$$s (n) = \frac{n+2-(-1)^n \left(n^2+4 n+2\right)}{(n+3) (n+4)}= \left\{ \begin {array} {ll} -\frac{n}{4+n} & n\;\;\text {even} \\ \frac{n+1}{n+3} & n\;\;\text {odd} \\ \end {array} \right. \tag{2}$$
Here is a plot of $s(n)$
it show that for large $n$ the sum oscillates between the borders $\pm 1$.
The proof of $(2)$ starts with observing that the inverse binomial coefficient is related to the beta function for which an integral representation is known:
$$\begin{align} \frac{k}{\binom{n+1}{k}}=\frac{k \times k!\times (-k+n+1)!}{(n+1)!}=\frac{k \times \Gamma (k+1) \Gamma (-k+n+2)}{\Gamma (n+2)}\\=\frac{k^2 \Gamma (k) \Gamma (-k+n+2)}{\Gamma (n+2)}=k^2 \int_0^1 x^{k-1} (1-x)^{-k+n+1} \, dx \end{align}\tag{3}$$
The sum to be calculated under the integral is
$$s_i=\sum _{k=1}^n (-1)^{k-1} k^2 x^{k-1} (1-x)^{-k+n+1}\\ =-\frac{1}{x} (1-x)^{n+1}\sum _{k=1}^n k^2 (\frac{-x}{1-x})^{k}\tag{4a}$$
It can be done using Feynman's trick on the geometric sum:
$$\sum _{k=1}^n k^2 z^k=(z \frac{\partial }{\partial z})(z \frac{\partial }{\partial z})\sum _{k=1}^n z^k= (z \frac{\partial }{\partial z})(z \frac{\partial }{\partial z})\frac{z^{n+1}-1}{z-1}\tag{4b}$$
which, after some careful simplications, gives
$$s_i=-(2 x-1) (1-x)^{n+2}-(-1)^n (1-x) \left(-(2 n+3) x+(n+1)^2+2 x^2\right) x^n\tag{4c}$$
The integral over $s_i$ can be written as $s:=i=i_1 + i_2$ where
$$\begin {align} i_1 & =-\int_0^1 (2 x-1) (1-x)^{n+2} \, dx\overset{1-x\to u} =-\int_0^1 (2 u-1) \left(u^{n+2}\right) \, du\\ & =\frac{2}{n+4}-\frac{1}{n+3} \end {align}\tag{5a}$$
$$\begin {align} i_2 & = \int_0^1 -(-1)^n (1-x) x^n \left(-(2 n+3) x+(n+1)^2+2 x^2\right) \, dx\\ & = (-1)^{n+1} \left(-\frac{2}{n+4}-\frac{1}{n+3}+1\right) \end {align}\tag{5b}$$
Adding the expressions in $(5a)$ and $(5b)$ proves $(2)$ which is easy to see in the formula with the parity alternatives.
The specific value to be calculated in the OP, is $s(19) = \frac{19+1}{19+3}=\frac{20}{22}=\frac{10}{11}$.
Discussion
With the same method (and with the help of a CAS which knows special functions) we can calculate the generating sum
$$\begin {align} g(n,z) & = \sum _{k=1}^n \frac{z^k}{\binom{n+1}{k}}=\int_0^1 \left(\sum _{k=1}^n x^{k-1} z^k (1-x)^{-k+n+1}\right) \, dx\\ & =\int_0^1 \frac{(x-1) z \left((1-x)^n-x^n z^n\right)}{x z+x-1} \, dx\\ & =\frac{1}{(n+1) (n+2)}(-(n+2) z^{n+1} \, _2F_1(1,n+1;n+2;z+1)\\ & +(n+1) z^{n+1} \, _2F_1(1,n+2;n+3;z+1)\\ & +(n+2) z \, _2F_1(1,1;n+2;z+1)\\ & -(n+1) \, _2F_1(1,1;n+3;z+1)+n+2) \end{align}\tag{a}$$
where $_2F_1(a,b;c;y)$ is the hypergeometric function. In the "physical" range $z\le 0$ these assume real values.
And we can find e.g.
$$\begin{align}s_2(n) & = \sum _{k=1}^n \frac{(-1)^{k-1} k^2}{\binom{n+1}{k}}\\ & = \frac{(n+1) \left(n+2-(-1)^n \left(n^3+8 n^2+16 n+2\right)\right)}{(n+3) (n+4) (n+5)}\\ & \underset{n \to \infty} \simeq (-1)^{n+1} \left(n-3+\frac{13}{n}\right)+\frac{1}{n}+O(\frac{1}{n^2})\end{align}\tag {b}$$
