Introduction:
The given function is a sum of an infinite series. We need to examine the continuity of this function at x=0.

Solution:
To check the continuity of the function at x=0, we need to check the left-hand limit, right-hand limit, and the value of the function at x=0.

Left-hand Limit:
Let us consider x approaching 0 from the negative side. As x approaches 0, all the terms in the series become negative. Therefore, we can write the sum of the series as:

(x/(x+1)) - (x/(x+1)(2x+1)) + (x/(2x+1)(3x+1)) - ...

As x approaches 0, the denominators of all the terms approach 1. Therefore, the limit of the sum of the series as x approaches 0 from the negative side can be written as:

lim x->0- [(x/(x+1)) - (x/(x+1)(2x+1)) + (x/(2x+1)(3x+1)) - ...]

= (0/1) - (0/1) + (0/1) - ...

= 0

Right-hand Limit:
Now, let us consider x approaching 0 from the positive side. As x approaches 0, all the terms in the series become positive. Therefore, we can write the sum of the series as:

(x/(x+1)) + (x/(x+1)(2x+1)) + (x/(2x+1)(3x+1)) + ...

As x approaches 0, the denominators of all the terms approach 1. Therefore, the limit of the sum of the series as x approaches 0 from the positive side can be written as:

lim x->0+ [(x/(x+1)) + (x/(x+1)(2x+1)) + (x/(2x+1)(3x+1)) + ...]

= (0/1) + (0/1) + (0/1) + ...

= 0

Value at x=0:
To find the value of the function at x=0, we substitute x=0 in the sum of the series. As all the terms have x in the denominator, the sum of the series is not defined at x=0.

Conclusion:
As the left-hand limit, right-hand limit, and the value of the function at x=0 are all equal to 0, the function is continuous at x=0. However, the sum of the series is not defined at x=0.