Prove the following equality using mathematical induction: $\sum\limits_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1}$

Question

Prove the following equality using mathematical induction: For any integer $n \ge 1$

$$\sum_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1}$$

I understand for the base base I need to have $n=1$. If I substituted $n$ for 1 and $i$ for 1 I would get

\begin{align} \sum_{i=1}^{1} \frac{1}{1(1+1)} &= \frac{1}{1+1} \\ \sum_{i=1}^{1} \frac{1}{2} &= \frac{1}{2} \end{align}

As for the inductive step, I inputted $n +1$ but it did not work for me because honestly I did not know what to do with the $i$ and what I should do next. How would I go about doing the next step and an explanation for $i$ would be helpful.

Answer 1

Hint $\: $ First trivially inductively prove the Fundamental Theorem of Difference Calculus

$$\rm\ F(n)\ =\ \sum_{i\:=\:1}^n\ f(i)\ \ \iff\ \ \ F(n\!+\!1) - F(n)\ =\ f(n),\quad\ F(1) =\: f(1)$$

Yours is $\rm\ F(n)-F(n\!-\!1)\ =\ n/(n\!+\!1)-(n\!-\!1)/n = 1/(n(n\!+\!1)).\,$ Note that by employing the Fundamental Theorem we have reduced the proof to the trivial verification of an equation. No ingenuity is required.

Note that the proof of the Fundamental Theorem is much more obvious than that for your special case because the telescopic cancellation is obvious at this level of generality, whereas it is usually obfuscated in most specific instances. For further discussion see my many posts on telescopy.

Answer 2

HINT:

If $\displaystyle\sum_{i=1}^m \frac1{i(i+1)} = \frac m{m+1}$

$\displaystyle\sum_{i=1}^{m+1} \frac1{i(i+1)}=\sum_{i=1}^m \frac1{i(i+1)}+\frac1{(m+1)(m+1+1)}$ $\displaystyle= \frac m{m+1}+\frac1{(m+1)(m+2)}=\frac1{m+1}\left(m+\frac1{m+2}\right)$

$\displaystyle =\frac1{m+1}\cdot\frac{(m+1)^2}{m+2}=\frac{m+1}{m+2}=\frac{(m+1)}{(m+1)+1}$

Answer 3

Hint:

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

And note that the sum on the left side is a telescop sum. Thus you do not even need induction to proof this.

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However, for the induction step, you could write

$$\begin{array}{rcl} \sum_{i=1}^{n+1} \frac{1}{i(i+1)} &=& \frac{1}{(n+1)(n+2)}+\sum_{i=1}^{n} \frac{1}{i(i+1)} \\ &=& 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} \\ &=& 1 - \frac{n+2}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)} \\ &=& 1 - \frac{n+2-1}{(n+1)(n+2)}\\ &=& 1 - \frac{1}{n+2}\\ \end{array}$$

Answer 4

Proving these kind of statements always goes the same way. In the induction step, you seperate the last summand $$\sum_{i=1}^{n+1} \frac 1{i(i+1)} = \sum_{i=1}^{n} \frac 1{i(i+1)} + \frac{1}{(n+1)(n+2)},$$ then you can apply the induction hypothesis to get $$\sum_{i=1}^{n+1} \frac 1{i(i+1)} = \frac{n}{n+1} + \frac{1}{(n+1)(n+2)}.$$ Put both fractions on a single denominator and obtain $$\sum_{i=1}^{n+1} \frac 1{i(i+1)} = \frac{n(n+2)+1}{(n+1)(n+2)} = \frac{n^2+2n+1}{(n+1)(n+2)} = \frac{(n+1)^2}{(n+1)(n+2)} = \frac{n+1}{n+2}. \square$$

Answer 5

The extra index does make it a little trickier. Be careful as you work through this problem and always remember that $n$ is the index for induction.

So the inductive hypothesis is that

$$ \sum_{i=1}^n \frac{1}{i(i + 1)} = \frac{n}{n+1} $$

and using this, you want to prove $$ \sum_{i=1}^{n+1} \frac{1}{i(i + 1)} = \frac{n+1}{(n+1)+1} = \frac{n+1}{n+2}. $$

To this end, you can write

$$ \sum_{i=1}^{n+1} \frac{1}{i(i + 1)} = \sum_{i=1}^{n} \frac{1}{i(i + 1)} + \frac{1}{(n+1)(n + 1 + 1)} $$ which by your induction hypothesis, is the same as $$ \frac{n}{n+1} + \frac{1}{(n+1)(n + 1 + 1)} $$

can you finish from here?