Let $$r_k = \frac{\int_0^1 (1-x^7)^k dx}{\int_0^1 (1-x^7)^{k+1} dx}$$, $$k \in \mathbb{N}$$. Then the value of $$\sum_{k=1}^{10} \frac{1}{7(r_k - 1)}$$ is equal to ________

Let us denote $$I_n = \displaystyle\int_{0}^{1} \bigl(1-x^{7}\bigr)^{n}\,dx$$ for every non-negative integer $$n$$. The given ratio can then be written as $$r_k = \dfrac{I_k}{I_{\,k+1}}$$.

Step 1: Evaluate $$I_n$$ in terms of the Beta function
Put $$x^{7}=t \;\Longrightarrow\; x = t^{1/7}, \; dx = \dfrac{1}{7}\,t^{-6/7}\,dt.$$ Then

$$I_n = \int_{0}^{1} (1-t)^{n}\,\dfrac{1}{7}\,t^{-6/7}\,dt = \dfrac{1}{7}\,B\!\left(\dfrac{1}{7},\,n+1\right)$$

Using $$B(p,q)=\dfrac{\Gamma(p)\,\Gamma(q)}{\Gamma(p+q)}$$, we get

$$I_n = \dfrac{1}{7}\, \dfrac{\Gamma\!\left(\dfrac{1}{7}\right)\,\Gamma(n+1)} {\Gamma\!\left(n+1+\dfrac{1}{7}\right)}.$$

Step 2: Simplify the ratio $$r_k$$
Write the expressions for two consecutive integrals:

$$I_k = \dfrac{1}{7}\, \dfrac{\Gamma\!\left(\dfrac{1}{7}\right)\,k!} {\Gamma\!\left(k+1+\dfrac{1}{7}\right)}, \qquad I_{k+1} = \dfrac{1}{7}\, \dfrac{\Gamma\!\left(\dfrac{1}{7}\right)\,(k+1)!} {\Gamma\!\left(k+2+\dfrac{1}{7}\right)}.$$

Therefore

$$r_k = \dfrac{I_k}{I_{k+1}} = \dfrac{k!}{(k+1)!}\; \dfrac{\Gamma\!\left(k+2+\dfrac{1}{7}\right)} {\Gamma\!\left(k+1+\dfrac{1}{7}\right)} = \dfrac{1}{k+1}\; \dfrac{\bigl(k+1+\tfrac17\bigr)\, \Gamma\!\left(k+1+\dfrac{1}{7}\right)} {\Gamma\!\left(k+1+\dfrac{1}{7}\right)} = \dfrac{k+1+\tfrac17}{k+1}.$$

Hence

$$r_k = 1 + \dfrac{1}{7(k+1)} \;\Longrightarrow\; r_k - 1 = \dfrac{1}{7(k+1)}.$$

Step 3: Evaluate the required sum

$$\frac{1}{7\,(r_k-1)} = \frac{1}{7}\; \frac{1}{\tfrac{1}{7(k+1)}} = k+1.$$

Thus

$$\sum_{k=1}^{10} \frac{1}{7\,(r_k-1)} = \sum_{k=1}^{10} (k+1) = 2 + 3 + \dots + 11.$$

This is an arithmetic series with 10 terms. Sum $$S = \dfrac{10\,(2+11)}{2} = 5 \times 13 = 65.$$

Therefore, the value of the given expression is $$65$$.