# Proof by induction $\frac1{1 \cdot 2} + \frac1{2 \cdot 3} + \frac1{3 \cdot 4} + \cdots + \frac1{n \cdot (n+1)} = \frac{n}{n+1}$

Need some help on following induction problem:

$$\dfrac1{1 \cdot 2} + \dfrac1{2 \cdot 3} + \dfrac1{3 \cdot 4} + \cdots + \dfrac1{n \cdot (n+1)} = \dfrac{n}{n+1}$$

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Every question of the form: prove by induction that $$\sum_{k=1}^n f(k)=g(n)$$ can be done by verifying two facts about the functions $f$ and $g$:

• $f(1)=g(1)$ and
• $g(n+1)-g(n)=f(n+1)$.
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Hint: $$\frac{1}{k(k+1)}= \frac{1}{k} - \frac{1}{k+1}.$$

Hint2:

$$\frac{n}{n+1} = 1 - \frac{1}{n+1}.$$

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$\textbf{HINT:}$ Actually this is the answer in itself.

$\frac{n}{n+1} + \frac{1}{(n+1)(n+2)} = \frac{n+1}{n+2}$.

This is all that you will need when you do induction.

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Essentially you test the base case, p(1), which is true since 1/(1+1) = 1/2. Then, you want to assume your p(n) is true, namely, that sum you wrote above:

1/2 + .... + 1/(n+1)


and then add the next number in the sequence.

1/[(n+1)+1]


Since you know your p(n) sum equals n/(n+1), you add the next number in the sequence above to your p(n):

1/2 + ... + 1/(n+1) + 1/[(n+1)+1] = p(n) + 1/[(n+1)+1] = n/(n+1) + 1/[(n+1)+1]

Given the useful hint above from user17762, they have done the algebra for you.

n/(n+1) + 1/[(n+1)+1] = (n+1)/(n+2)


You can rewrite as, (n+1)/(n+1+1), which is exactly p(n+1), so we assume it's true for all n because it's true for the next term in the sequence. B.Y.U.

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