Let $a_{n}$ be the area surrounded by the curves $y=e^{-x}$ and the part of $y=e^{-x}|\cos x|,\ (n-1)\pi \leq x\leq n\pi \ (n=1,\ 2,\ 3,\ \cdots).$ 
Evaluate $\lim_{n\to\infty}(a_{1}+a_{2}+\cdots+a_{n}).$
1. First, we need to find the area \(a_n\) between the curves \(y = e^{-x}\) and \(y = e^{-x}|\cos x|\) over the interval \((n-1)\pi \leq x \leq n\pi\). This can be expressed as:
   \[
   a_n = \int_{(n-1)\pi}^{n\pi} \left(e^{-x} - e^{-x}|\cos x|\right) \, dx = \int_{(n-1)\pi}^{n\pi} e^{-x}(1 - |\cos x|) \, dx
   \]

2. To simplify the integral, we perform a change of variables. Let \(x = (n-1)\pi + y\), where \(0 \leq y \leq \pi\). Then \(dx = dy\), and the integral becomes:
   \[
   a_n = e^{-(n-1)\pi} \int_0^\pi e^{-y}(1 - |\cos y|) \, dy
   \]

3. Next, we need to evaluate the integral:
   \[
   \int_0^\pi e^{-y}(1 - |\cos y|) \, dy
   \]
   This can be split into two integrals:
   \[
   \int_0^\pi e^{-y}(1 - |\cos y|) \, dy = \int_0^\pi e^{-y} \, dy - \int_0^\pi e^{-y}|\cos y| \, dy
   \]

4. The first integral is straightforward:
   \[
   \int_0^\pi e^{-y} \, dy = \left[-e^{-y}\right]_0^\pi = 1 - e^{-\pi}
   \]

5. For the second integral, we split it into two parts due to the absolute value:
   \[
   \int_0^\pi e^{-y}|\cos y| \, dy = \int_0^{\pi/2} e^{-y}\cos y \, dy + \int_{\pi/2}^\pi e^{-y}(-\cos y) \, dy
   \]

6. Using integration by parts, we find:
   \[
   \int e^{-y}\cos y \, dy = e^{-y}\sin y + \int e^{-y}\sin y \, dy
   \]
   \[
   \int e^{-y}\sin y \, dy = -e^{-y}\cos y - \int e^{-y}\cos y \, dy
   \]
   Solving these, we get:
   \[
   \int e^{-y}\cos y \, dy = \frac{1}{2} e^{-y}(\sin y - \cos y)
   \]

7. Evaluating the definite integrals:
   \[
   \int_0^{\pi/2} e^{-y}\cos y \, dy = \left[\frac{1}{2} e^{-y}(\sin y - \cos y)\right]_0^{\pi/2} = \frac{1}{2}(e^{-\pi/2} - 1)
   \]
   \[
   \int_{\pi/2}^\pi e^{-y}(-\cos y) \, dy = \left[-\frac{1}{2} e^{-y}(\sin y - \cos y)\right]_{\pi/2}^\pi = \frac{1}{2}(e^{-\pi} - e^{-\pi/2})
   \]

8. Combining these results:
   \[
   \int_0^\pi e^{-y}|\cos y| \, dy = \frac{1}{2}(e^{-\pi/2} - 1) + \frac{1}{2}(e^{-\pi} - e^{-\pi/2}) = \frac{1}{2}(1 - e^{-\pi})
   \]

9. Therefore:
   \[
   \int_0^\pi e^{-y}(1 - |\cos y|) \, dy = (1 - e^{-\pi}) - \frac{1}{2}(1 - e^{-\pi}) = \frac{1}{2}(1 - e^{-\pi})
   \]

10. Substituting back, we get:
    \[
    a_n = e^{-(n-1)\pi} \cdot \frac{1}{2}(1 - e^{-\pi}) = \frac{1}{2} e^{-(n-1)\pi} (1 - e^{-\pi})
    \]

11. Summing \(a_n\) from \(n = 1\) to \(\infty\):
    \[
    \sum_{n=1}^\infty a_n = \frac{1}{2} (1 - e^{-\pi}) \sum_{n=1}^\infty e^{-(n-1)\pi}
    \]

12. The series \(\sum_{n=1}^\infty e^{-(n-1)\pi}\) is a geometric series with the first term \(a = 1\) and common ratio \(r = e^{-\pi}\):
    \[
    \sum_{n=1}^\infty e^{-(n-1)\pi} = \frac{1}{1 - e^{-\pi}}
    \]

13. Therefore:
    \[
    \sum_{n=1}^\infty a_n = \frac{1}{2} (1 - e^{-\pi}) \cdot \frac{1}{1 - e^{-\pi}} = \frac{1}{2}
    \]

The final answer is \(\boxed{\frac{1}{2}}\)