1. **Problem:** Find the value of $a$ such that $$\lim_{x \to +\infty} \left(\frac{x+a+1}{x+1}\right)^{x-1} = \frac{1}{\sqrt{e}}.$$ 2. **Formula and rules:** Recall the limit definition of the exponential function: $$\lim_{x \to \infty} \left(1 + \frac{k}{x}\right)^x = e^k.$$ We want to rewrite the expression inside the limit in a form similar to $\left(1 + \frac{k}{x}\right)^x$. 3. **Rewrite the base:** $$\frac{x+a+1}{x+1} = \frac{x+1 + a}{x+1} = 1 + \frac{a}{x+1}.$$ 4. **Rewrite the limit:** $$\lim_{x \to +\infty} \left(1 + \frac{a}{x+1}\right)^{x-1} = \lim_{x \to +\infty} \left[\left(1 + \frac{a}{x+1}\right)^{x+1}\right]^{\frac{x-1}{x+1}}.$$ 5. **Evaluate the inner limit:** As $x \to +\infty$, $$\left(1 + \frac{a}{x+1}\right)^{x+1} \to e^a.$$ 6. **Evaluate the exponent:** $$\frac{x-1}{x+1} \to 1.$$ 7. **Combine results:** $$\lim_{x \to +\infty} \left(\frac{x+a+1}{x+1}\right)^{x-1} = e^a.$$ 8. **Set equal to given limit:** $$e^a = \frac{1}{\sqrt{e}} = e^{-\frac{1}{2}}.$$ 9. **Solve for $a$:** $$a = -\frac{1}{2}.$$ **Final answer:** $$\boxed{a = -\frac{1}{2}}.$$