Mathematical induction and sequences $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+\dots+n}=\frac{2n}{n+1}$

Question

Prove the following sequence with the help of the principle of mathematical induction

$$1+\frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+\dots+n}=\frac{2n}{n+1}$$

Answer 1

Hint:

$\sum\limits_{i=1}^n i = \frac{n^2+n}{2}$ so the original sum can be rewritten as:

$$1+\frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+\dots+n}=\sum\limits_{i=1}^n \frac{2}{i^2+i}$$

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Set up the induction proof as normal. For tips on how to format/setup an induction proof, I recommend reading this question and answer. During the induction step, you should be attempting to prove that $\sum\limits_{i=1}^{n+1}\frac{2}{i^2+i}=\frac{2(n+1)}{(n+1)+1}$ by using your induction hypothesis and algebraic manipulation. Now, note that:

$$\sum\limits_{i=1}^{n+1}\frac{2}{i^2+i}=\frac{2}{(n+1)^2+(n+1)}+\sum\limits_{i=1}^n\frac{2}{i^2+i}$$

You may now apply your induction hypothesis. Try then to add the two and algebraically manipulate the result to look like the intended outcome.

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For an even more direct simplification which lends to a direct proof, one can note further that $\frac{2}{i^2+i}=\frac{2}{i}-\frac{2}{i+1}$, so the series telescopes, giving rather immediately that $\sum\limits_{i=1}^n \frac{2}{i^2+i}=\frac{2}{1}-\frac{2}{n+1}$