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Find a formula for a sum of 1/k(k+1)

by RoRi

Claim:

    \[ \sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}\]

Proof. The proof is by induction. If n = 1 then on the left we have

    \[ \sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^1 \frac{1}{k(k+1)} = \frac{1}{2}. \]

On the right,

    \[ 1 - \frac{1}{n+1} = 1 - \frac{1}{2} = \frac{1}{2}.\]

So, indeed, the formula holds for the case n=1.
Assume then that the formula is true for some n=m \in \mathbb{Z}_{\geq 1}. Then,

    \begin{align*} &&\sum_{k=1}^m \frac{1}{k(k+1)} &= 1 - \frac{1}{m+1} \\ \implies && \left(\sum_{k=1}^{m} \frac{1}{k(k+1)}\right) + \frac{1}{(m+1)(m+2)} &= 1 - \frac{1}{m+1} + \frac{1}{(m+1)(m+2)} \\ \implies && \sum_{k=1}^{m+1} \frac{1}{k(k+1)} &= 1 - \frac{m+2-1}{(m+1)(m+2)}\\ \implies && \sum_{k=1}^{m+1} \frac{1}{k(k+1)} &= 1 - \frac{1}{m+2} \end{align*}

Thus, the formula holds for m+1; and thus, for all n \in \mathbb{Z}_{\geq 1}. \qquad \blacksquare

3 comments

  3. LAURA NATALIA MERCHAN OLAYA says:

    Buenos dias, quisiera saber porque al hacer la suma de fraccionarios, da n+2-1 y no n+2+1, ya que es una suma de fraccionarios y no una resta, gracias.

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