Limit n tends to infinite [1/6+1/24+1/60+1/120+...... +1/(n³-n) ]

To evaluate the limit as \( n \) approaches infinity for the series \( \frac{1}{6} + \frac{1}{24} + \frac{1}{60} + \frac{1}{120} + \ldots + \frac{1}{n^3 - n} \), we first need to analyze the general term of the series, which is \( \frac{1}{n^3 - n} \). This term can be simplified and understood better to find the limit.

Breaking Down the General Term

The expression \( n^3 - n \) can be factored as follows:

• Factor out \( n \): \( n(n^2 - 1) \)
• Further factor \( n^2 - 1 \): \( n(n - 1)(n + 1) \)

Thus, we can rewrite the general term:

General Term: \( \frac{1}{n^3 - n} = \frac{1}{n(n - 1)(n + 1)} \)

Understanding the Behavior as n Approaches Infinity

As \( n \) becomes very large, the term \( n(n - 1)(n + 1) \) grows significantly. Specifically, we can approximate it as:

Approximation: \( n(n - 1)(n + 1) \approx n^3 \)

This leads us to the conclusion that:

Limit of the General Term: \( \frac{1}{n^3 - n} \approx \frac{1}{n^3} \) as \( n \to \infty \)

Summing the Series

Now, we need to consider the sum of these terms from \( 1 \) to \( n \). The series can be expressed as:

Series: \( \sum_{k=1}^{n} \frac{1}{k^3 - k} \approx \sum_{k=1}^{n} \frac{1}{k^3} \)

We know that the series \( \sum_{k=1}^{n} \frac{1}{k^3} \) converges to a finite value as \( n \) approaches infinity. In fact, it converges to \( \frac{\pi^2}{6} \) for the infinite series, but we are interested in the behavior of the terms as \( n \) increases.

Evaluating the Limit

To find the limit of the entire sum as \( n \) approaches infinity, we can use the fact that:

Limit of the Series: \( \sum_{k=1}^{\infty} \frac{1}{k^3} \) converges, and thus:

As \( n \to \infty, \sum_{k=1}^{n} \frac{1}{k^3 - k} \to \text{finite value} \approx \sum_{k=1}^{\infty} \frac{1}{k^3} \)

Therefore, the limit of the series \( \frac{1}{6} + \frac{1}{24} + \frac{1}{60} + \frac{1}{120} + \ldots + \frac{1}{n^3 - n} \) as \( n \) approaches infinity converges to a finite value.

Final Thoughts

In summary, the limit of the series you provided converges to a finite value as \( n \) approaches infinity. The terms decrease rapidly enough that their sum does not diverge, and we can conclude that:

Limit: \( \lim_{n \to \infty} \left( \frac{1}{6} + \frac{1}{24} + \frac{1}{60} + \ldots + \frac{1}{n^3 - n} \right) \) converges to a finite number.