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Peter
Peter
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// n is the numerator, d the denominator, i the amount of numbers that have been added and j the denominator of the fraction that is supposed to be added.
n,d,i,j;
s(a)
{
    // The update statement of the for loop is moved to the end of the loop body for clearness (it's executed at the end of the loop anyways).
    // a / i is equivalent to a >= i, so we loop until i is greater than a
    for(n = i = 1, d = j = 2; a / i;)
        // Both n and the d isare multiplied by j, and the previous value of d is added to n.
        // The new fraction is (n*j + d) / (d*j).
        // (n*j)/(d*j) = n/d, so what's actually added to the fraction is d/(d*j), which can be rewritten as 1/j.
        n = n * ++j + d,
        d *= j,

        // n / d is non-zero iff n >= d, and n >= d iff n/d >= 1
        // If n/d >= 1, subtract what was previously added to n, twice.
        // Otherwise, increment i
        n / d ? n -= 2 * d / j : ++i;
    return j;
}

// n is the numerator, d the denominator, i the amount of numbers that have been added and j the denominator of the fraction that is supposed to be added.
n,d,i,j;
s(a)
{
    // The update statement of the for loop is moved to the end of the loop body for clearness (it's executed at the end of the loop anyways).
    // a / i is equivalent to a >= i, so we loop until i is greater than a
    for(n = i = 1, d = j = 2; a / i;)
        // Both n and the d is multiplied by j, and the previous value of d is added to n.
        // The new fraction is (n*j + d) / (d*j).
        // (n*j)/(d*j) = n/d, so what's actually added to the fraction is d/(d*j), which can be rewritten as 1/j.
        n = n * ++j + d,
        d *= j,

        // n / d is non-zero iff n >= d, and n >= d iff n/d >= 1
        // If n/d >= 1, subtract what was previously added to n, twice.
        // Otherwise, increment i
        n / d ? n -= 2 * d / j : ++i;
    return j;
}

// n is the numerator, d the denominator, i the amount of numbers that have been added and j the denominator of the fraction that is supposed to be added.
n,d,i,j;
s(a)
{
    // The update statement of the for loop is moved to the end of the loop body for clearness (it's executed at the end of the loop anyways).
    // a / i is equivalent to a >= i, so we loop until i is greater than a
    for(n = i = 1, d = j = 2; a / i;)
        // Both n and d are multiplied by j, and the previous value of d is added to n.
        // The new fraction is (n*j + d) / (d*j).
        // (n*j)/(d*j) = n/d, so what's actually added to the fraction is d/(d*j), which can be rewritten as 1/j.
        n = n * ++j + d,
        d *= j,

        // n / d is non-zero iff n >= d, and n >= d iff n/d >= 1
        // If n/d >= 1, subtract what was previously added to n, twice.
        // Otherwise, increment i
        n / d ? n -= 2 * d / j : ++i;
    return j;
}
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Peter
Peter
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This 70-byte version also seems to work, although I'm not sure if it always does:

n,d,i,j;s(a){for(n=i=1,d=j=2;a/i;n/d?n-=2*d/j:++i)n=n*++j+d,d*=j;a=j;}

The 70-byte version stores j in a, which seems to have the same effect as returning j. If anyone knows, please tell me if that behavior is consistent and I'm allowed to use it.

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This 70-byte version also seems to work, although I'm not sure if it always does:

n,d,i,j;s(a){for(n=i=1,d=j=2;a/i;n/d?n-=2*d/j:++i)n=n*++j+d,d*=j;a=j;}

The 70-byte version stores j in a, which seems to have the same effect as returning j. If anyone knows, please tell me if that behavior is consistent and I'm allowed to use it.

Try It Online!

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Peter
Peter
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C (GCC), 75 bytes

n,d,i,j;s(a){for(n=i=1,d=j=2;a/i;n/d?n-=2*d/j:++i)n=n*++j+d,d*=j;return j;}

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Overflows when the input is greater than 8.

How it works:

// n is the numerator, d the denominator, i the amount of numbers that have been added and j the denominator of the fraction that is supposed to be added.
n,d,i,j;
s(a)
{
    // The update statement of the for loop is moved to the end of the loop body for clearness (it's executed at the end of the loop anyways).
    // a / i is equivalent to a >= i, so we loop until i is greater than a
    for(n = i = 1, d = j = 2; a / i;)
        // Both n and the d is multiplied by j, and the previous value of d is added to n.
        // The new fraction is (n*j + d) / (d*j).
        // (n*j)/(d*j) = n/d, so what's actually added to the fraction is d/(d*j), which can be rewritten as 1/j.
        n = n * ++j + d,
        d *= j,

        // n / d is non-zero iff n >= d, and n >= d iff n/d >= 1
        // If n/d >= 1, subtract what was previously added to n, twice.
        // Otherwise, increment i
        n / d ? n -= 2 * d / j : ++i;
    return j;
}

Here's a 127-byte version that overflows at a number of terms somewhere between 21 and 57:

long long n,d,i,j,k;s(a){for(n=i=1,d=j=2;a/i;n/d?n-=2*d/j:++i){n=n*++j+d,d*=j;for(k=1;k<99;)n%++k||d%k||(n/=k,d/=k);}return j;}

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It's pretty much the same as the first one, except the variables are long long integers rather than integers, and it includes for(k=1;k<99;)n%++k||d%k||(n/=k,d/=k);, which, for each number k from 1 to 98, divides n and d by k if they're both divisible by k.