Final answer:
To prove the inequality 1/2² + 1/3² + 1/4² + ... + 1/n² < n-1/n, we can use the concept of telescoping series. By factoring out (1/n) from each term, we can rewrite the left-hand side of the inequality as (1/2) + (1/3) + (1/4) + ... + (1/n). Using the fact that the harmonic series diverges, we can conclude that this sum is always less than 1 and thus less than 1 - (1/n), proving the inequality.
Step-by-step explanation:
To prove the inequality 1/2² + 1/3² + 1/4² + ... + 1/n² < n-1/n, we can use the concept of telescoping series.
Let's start with the left-hand side of the inequality. We can rewrite each term as 1/n² = (1/n) * (1/n).
Now, let's consider the sum of the first n terms on the left-hand side: 1/2² + 1/3² + 1/4² + ... + 1/n² = (1/2) * (1/2) + (1/3) * (1/3) + (1/4) * (1/4) + ... + (1/n) * (1/n).
By factoring out (1/n) from each term, we get
((1/2) * (1/2)) + ((1/3) * (1/3)) + ((1/4) * (1/4)) + ... + ((1/n) * (1/n)) = (1/2) + (1/3) + (1/4) + ... + (1/n).
Now, let's compare this sum to the right-hand side of the inequality. The right-hand side is n-1/n, which can be written as (n/n) - (1/n) = 1 - (1/n).
To prove that the sum of the first n terms is less than 1 - (1/n), we can use the fact that the harmonic series, which is the sum of the reciprocals of positive integers (1 + 1/2 + 1/3 + ... + 1/n), is known to diverge.
Since the harmonic series diverges and our sum is equal to a subsequence of the harmonic series, our sum is also divergent. This means that our sum is always less than the value of the harmonic series, which is greater than 1.
Therefore, we have proved that 1/2² + 1/3² + 1/4² + ... + 1/n² < n-1/n.