The task presented in the image is asking to prove the following sum by induction:

$\sum_{\nu=1}^{n-1} \frac{1}{\nu (\nu + 1)} = \frac{n-1}{n} \quad \text{for} \quad n \geq 2$

This problem requires the use of mathematical induction to prove the given identity. Here’s a step-by-step outline of how we can approach this:

1. Base Case (n = 2):

We first check the base case where $n = 2$. $\sum_{\nu=1}^{2-1} \frac{1}{\nu (\nu + 1)} = \sum_{\nu=1}^{1} \frac{1}{1(1+1)} = \frac{1}{2}.$ Now check the right-hand side of the equation: $\frac{2-1}{2} = \frac{1}{2}.$ Since both sides are equal, the base case holds.

2. Inductive Step:

Assume that the formula holds for some $n = k$. This means that we assume: $\sum_{\nu=1}^{k-1} \frac{1}{\nu (\nu + 1)} = \frac{k-1}{k}.$ Now, we need to prove that the formula holds for $n = k+1$, i.e., $\sum_{\nu=1}^{k} \frac{1}{\nu (\nu + 1)} = \frac{k}{k+1}.$ Start by using the inductive hypothesis: $\sum_{\nu=1}^{k} \frac{1}{\nu (\nu + 1)} = \sum_{\nu=1}^{k-1} \frac{1}{\nu (\nu + 1)} + \frac{1}{k(k+1)}.$ From the inductive hypothesis, we substitute the first sum: $\frac{k-1}{k} + \frac{1}{k(k+1)}.$ Now simplify the right-hand side: $\frac{k-1}{k} + \frac{1}{k(k+1)} = \frac{(k-1)(k+1) + 1}{k(k+1)} = \frac{k^2 - 1 + 1}{k(k+1)} = \frac{k^2}{k(k+1)} = \frac{k}{k+1}.$ Thus, we have proven the inductive step.

Conclusion:

Since the base case holds and the inductive step is valid, by the principle of mathematical induction, the formula is true for all $n \geq 2$.

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Related Questions:

1. How would you prove an alternative formula involving summations like this?
2. Can we use induction to prove summations involving higher powers, such as $\sum \frac{1}{\nu^2}$?
3. How does the structure of a series affect the induction step?
4. What other summation formulas can be derived using induction?
5. How does induction apply to more complex mathematical identities?

Tip:

When using induction, always ensure that the base case is checked carefully, as it forms the foundation for the rest of the proof.