Question

$\sum_{\gamma=0}^{n} \frac{(-1)^{\gamma}}{^{n}C_{\gamma}}$

Answer 1

$\begin{cases} 0,& \text{if } n \text{ is odd}, \\ \frac{2(n+1)}{n+2}, & \text{if } n \text{ is even}. \end{cases}$

Answer 2

We want to evaluate

$S(n)=\sum_{\gamma=0}^{n}\frac{(-1)^{\gamma}}{\binom{n}{\gamma}}$.

Step 1. Rewrite the summation:

Recall that

$\binom{n}{\gamma}=\frac{n!}{\gamma!(n-\gamma)!}\quad\Longrightarrow\quad \frac{1}{\binom{n}{\gamma}}=\frac{\gamma!(n-\gamma)!}{n!}$.

Thus,

$S(n)=\sum_{\gamma=0}^{n}(-1)^{\gamma}\frac{\gamma!(n-\gamma)!}{n!}$.

Step 2. Convert into an Integral Using the Beta Function:

Notice that

$\int_0^1 x^{\gamma}(1-x)^{n-\gamma}\,dx=\frac{\gamma!(n-\gamma)!}{(n+1)!}$.

This implies

$\frac{\gamma!(n-\gamma)!}{n!}=(n+1)\int_0^1 x^{\gamma}(1-x)^{n-\gamma}\,dx$.

Thus, we can write

$S(n)=(n+1)\int_0^1\left[\sum_{\gamma=0}^{n}(-1)^{\gamma}x^{\gamma}(1-x)^{n-\gamma}\right]dx$.

Step 3. Evaluate the Sum inside the Integral:

Note that

$(1-x)^{n-\gamma}x^{\gamma}=(1-x)^n\left(\frac{x}{1-x}\right)^{\gamma}$.

So the inner summation becomes

$\sum_{\gamma=0}^{n}(-1)^{\gamma}\left(\frac{x}{1-x}\right)^{\gamma}=(1-x) \cdot \frac{1- \left(-\frac{x}{1-x}\right)^{n+1}}{1-x+x}$

since

$1+\frac{x}{1-x}=\frac{1}{1-x}$.

In fact, a simpler way is to recognize it as a finite geometric series: Let

$r=-\frac{x}{1-x}$.

Then

$\sum_{\gamma=0}^{n}r^{\gamma}=\frac{1-r^{n+1}}{1-r}$.

Substituting back, we have

$\sum_{\gamma=0}^{n}(-1)^{\gamma}\left(\frac{x}{1-x}\right)^{\gamma}=\frac{1-\left(-\frac{x}{1-x}\right)^{n+1}}{1+\frac{x}{1-x}}=(1-x)\left[1-(-1)^{n+1}\left(\frac{x}{1-x}\right)^{n+1}\right]$.

Thus the inner expression becomes

$(1-x)^n \cdot (1-x)\left[1-(-1)^{n+1}\left(\frac{x}{1-x}\right)^{n+1}\right] = (1-x)^{n+1}\left[1-(-1)^{n+1}\frac{x^{n+1}}{(1-x)^{n+1}}\right]$.

That is,

$(1-x)^{n+1}-(-1)^{n+1}x^{n+1}$.

Step 4. Complete the Integral:

So

$S(n)=(n+1)\int_{0}^{1}\left[(1-x)^{n+1}-(-1)^{n+1}x^{n+1}\right]dx$.

We now evaluate the integrals separately:

1. $\displaystyle \int_0^1 (1-x)^{n+1}\,dx=\frac{1}{n+2}.$

2. $\displaystyle \int_0^1 x^{n+1}\,dx=\frac{1}{n+2}.$

Thus,

$S(n)=(n+1)\left[\frac{1}{n+2}-(-1)^{n+1}\frac{1}{n+2}\right]=\frac{n+1}{n+2}\left[1-(-1)^{n+1}\right]$.

Step 5. Write the Final Answer According to the Parity of n:

• If $n$ is odd then $n+1$ is even so $(-1)^{n+1}=1$ and

$S(n)=\frac{n+1}{n+2}[1-1]=0$.

• If $n$ is even then $n+1$ is odd so $(-1)^{n+1}=-1$ and

$S(n)=\frac{n+1}{n+2}[1-(-1)]=\frac{2(n+1)}{n+2}$.