Prove $\sum\limits_{i=1}^n \frac{1}{i(i+1)} = \frac{n}{n+1}$ by induction

Question

Using induction, prove that

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

Any help would be appreciated.

Note that 1/1*2+1/2*3+... is interpreted as $\frac{1}{1}\cdot 2 + \frac{1}{1}\cdot 3 + \dots$. You presumably mean instead 1/(1*2) + 1/(2*3)+... which is $\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3}+\dots$. When typing without proper typesetting, parenthesis are important. Visit this page for a primer on how to type in $\LaTeX$ and MathJax on this site in order to make your equations easier to read and make sure that there is no potential confusion in meaning of equations. — JMoravitz, Oct 29 '15 at 04:37
Possible duplicate of Prove the following equality using mathematical induction: — N. F. Taussig, Oct 29 '15 at 10:06

score 2 · Answer 1 · answered Oct 29 '15 at 04:52

Assume induction statement

\begin{equation} P(n): \sum\limits_{i=1}^n \frac{1}{i(i+1)} = \frac{n}{n+1} \end{equation}

For $n=1$,

\begin{equation} \sum\limits_{i=1}^1 \frac{1}{i(i+1)} = \frac{1}{1(1+1)} = \frac {1}{1+1} \end{equation}

Then assume $P(n)$ is true and $n \geq 1$. We must show the following induction statement to be true

\begin{equation} P(n+1): \sum\limits_{i=1}^{n+1} \frac{1}{i(i+1)} = \frac{n+1}{(n+1)+1} \end{equation}

Since we are assuming $P(n)$ is true,

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

Thus $P(n+1)$ is true so $\sum\limits_{i=1}^n \frac{1}{i(i+1)} = \frac{n}{n+1}$ is true for every $n \in \mathbb{N}$

score 1 · Answer 2 · answered Oct 29 '15 at 04:40

1

Note that $1\over (n+1)(n+2)$$=$${1\over n+1}-{1\over n+2}$. The result should immediately follows from this and your induction hypothesis.

answered Oct 29 '15 at 04:40

cr001

12,598

score 1 · Answer 3 · answered Oct 29 '15 at 04:50

To show: $$ \sum_{k=1}^n \frac 1 {k (k+1)} = \frac n {n+1} \tag{*} $$ Base case $n = 1$: obvious.

Assume (*) is true for $n$. Then $$ \begin{align} \sum_{k=1}^{n+1} \frac 1 {k (k+1)} &= \sum_{k=1}^n \frac 1 {k (k+1)} + \frac 1 {(n+1) (n+2)} \\ &= \frac n {n+1} + \frac 1 {(n+1) (n+2)} \tag{by IH} \\ &= \frac {n(n+2) + 1} {(n+1) (n+2)} \\ &= \frac {n^2+2n + 1} {(n+1) (n+2)} \\ &= \frac {(n + 1)^2} {(n+1) (n+2)} \\ &= \frac {(n + 1)} {(n+2)} \\ \end{align} $$ which shows that (*) is true for $n+1$.

score 1 · Answer 4 · answered Oct 29 '15 at 04:59

Base Case:

Suppose $n = 1$. Then,

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

Thus, the proposition holds for $n = 1$.

Inductive Step:

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

For arbitrary $n \in \mathbb{N}$.
Then we can deduce the following at $n + 1$:

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

Thus, $ \sum_{i = 1}^n \frac{1}{i(i + 1)} = \frac{n}{n + 1} $ holds for all $n \in \mathbb{N}$.

Prove $\sum\limits_{i=1}^n \frac{1}{i(i+1)} = \frac{n}{n+1}$ by induction

4 Answers4