Question

Show that: ∑(n goes from 0 to ∞)1/((α+n)(α+n+1)) =1/α , for α>0.

Answer 1

Answer on Question #51427 - Math - Real Analysis

Show that:

$$\sum_{n=0}^{\infty} \frac{1}{(a + n) * (a + n + 1)} = \frac{1}{a} \quad \text{for } a > 0;$$

Solution

$$\begin{aligned} & \frac{1}{(a + n) * (a + n + 1)} = \frac{1 + a + n - a - n}{(a + n) * (a + n + 1)} = \frac{(1 + a + n) - (a + n)}{(a + n) * (a + n + 1)} = \\ & = \frac{1 + a + n}{(a + n) * (a + n + 1)} - \frac{a + n}{(a + n) * (a + n + 1)} = \frac{1}{a + n} - \frac{1}{a + n + 1}; \end{aligned}$$

Let

$$\begin{aligned} S_N = \sum_{n=0}^{N} \frac{1}{(a + n) * (a + n + 1)} = \sum_{n=0}^{N} \left(\frac{1}{a + n} - \frac{1}{a + n + 1}\right) &= \left(\frac{1}{a} - \frac{1}{a + 1}\right) + \left(\frac{1}{a + 1} - \frac{1}{a + 2}\right) + \cdots + \\ &+ \left(\frac{1}{a + N} - \frac{1}{a + N + 1}\right) = \frac{1}{a} - \frac{1}{a + 1} + \frac{1}{a + 1} - \frac{1}{a + 2} \cdots + \frac{1}{a + N} - \frac{1}{a + N + 1} = \frac{1}{a} - \frac{1}{a + N + 1}; \end{aligned}$$ $$S = \sum_{n=0}^{\infty} \frac{1}{(a + n)(a + n + 1)} = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \left(\frac{1}{a} - \frac{1}{a + N + 1}\right) = \frac{1}{a} - 0 = \frac{1}{a}$$

Answer: $\frac{1}{a}$.

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