Sum of Triangle Areas

For an integer parameter k, define S_k as the area of the triangle formed by the x-axis and the lines

\(l_{k1}:y=kx+k-1\) and \(l_{k2}:y=(k+1)x+k\)

and their point of intersection. Note that the intersection point of \(l_{k1}\) and \(l_{k2}\) is always \((-1,-1)\).

The other two vertices are obtained by the intersection of each line with the x-axis:

• For \(l_{k1}\): set \(kx+k-1=0\) to obtain \(x=\frac{1}{k}-1\) (provided \(k\ne0\)).
• For \(l_{k2}\): set \((k+1)x+k=0\) to obtain \(x=-\frac{k}{k+1}\) (provided \(k\ne-1\)).

Using the three vertices, the area of the triangle can be computed as

\(S_k=\frac{1}{2}\Big|\frac{1}{k} -1+\frac{k}{k+1}\Big|\).

Simplifying the expression, we find that

\(\frac{1}{k} -1 +\frac{k}{k+1}= \frac{1}{k(k+1)}\), so that

\(S_k=\frac{1}{2k(k+1)}\).

For a given positive integer n, we define the sum

\(\sum_{i=1}^{n} S_i = \frac{1}{2}\sum_{i=1}^{n} \frac{1}{i(i+1)}\).

Notice that the above sum telescopes and the closed form is

\(\frac{1}{2}\Big(1-\frac{1}{n+1}\Big)\) for \(n\ge1\). For \(n<1\), the sum is defined to be 0.

There are t test cases. In each test case, you are given an integer \(n_i\) and are required to output the value of the sum for \(n=n_i\).

inputFormat

The input begins with an integer t representing the number of test cases. Each of the following t lines contains an integer \(n_i\).

Note: If \(n_i < 1\), the sum is defined as 0.

outputFormat

For each test case, output a single line containing the value of \( \frac{1}{2}\Big(1-\frac{1}{n_i+1}\Big)\) if \(n_i\ge1\), or 0 if \(n_i < 1\). The answer should be printed as a floating-point number with 6 decimal places.

sample

3
1
2
3

0.250000
0.333333
0.375000

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