Chứng minh rằng 1/6 + 1/7 + ... + 1/36 > 1

To solve this problem, we can find a lower bound for the sum of the given series.

First, let's consider the first term of the series, 1/6. We can rewrite it as 1/(2*3).

Next, let's consider the second term of the series, 1/7. We can rewrite it as 1/(2*4).

Continuing this pattern, we can rewrite the nth term of the series as 1/(2*(n+4)).

Now, let's rewrite the sum of the series using these expressions:

1/6 + 1/7 + ... + 1/36 = 1/(2*3) + 1/(2*4) + ... + 1/(2*(36+4))

= 1/2 * (1/3 + 1/4 + ... + 1/40)

Since each term in the parentheses is positive, we can see that the sum of the series is greater than 1/2 times the sum of the series 1/3 + 1/4 + ... + 1/40.

Now, let's consider the sum of the series 1/3 + 1/4 + ... + 1/40. We can rewrite it as:

1/3 + 1/4 + ... + 1/40 = 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/40

= (1/3 + 1/4) + (1/5 + 1/6) + ... + (1/39 + 1/40)

= (7/12) + (11/30) + ... + (79/1560)

Each term in the parentheses is greater than 1/2, so the sum of the series 1/3 + 1/4 + ... + 1/40 is greater than (1/2) * (1/2 + 1/2 + ... + 1/2) = (1/2) * (39/2) = 39/4.

Therefore, the sum of the series 1/6 + 1/7 + ... + 1/36 is greater than 1/2 * (39/4) = 39/8.

Since 39/8 is greater than 1, we can conclude that:

1/6 + 1/7 + ... + 1/36 > 1.