# Sigma Notation

Sigma Notation is a practice to express the sum of a lengthy series in a simple and concise way. The notation used for this is $\Sigma$ and it is always followed by the variable we are summing over. This symbol tells us to add up everything that follows it. The numbers we add together are called the terms of the series.

##### 2017 EJC Promo Q7 (b) [Modified]

Given that $\sum\limits_{r=2}^{n}{\frac{1}{{{r}^{2}}-1}}=\frac{3}{4}-\frac{1}{2n}-\frac{1}{2\left( n+1 \right)}$,

(ii)

state $\sum\limits_{r=2}^{\infty }{\frac{1}{{{r}^{2}}-1}}$,

(ii) state $\sum\limits_{r=2}^{\infty }{\frac{1}{{{r}^{2}}-1}}$,

(iii)

find $\sum\limits_{r=2}^{n-1}{\frac{1}{r(r+2)}}$.

(iii) find $\sum\limits_{r=2}^{n-1}{\frac{1}{r(r+2)}}$.

##### 2018 YJC Promo Q13

The terms of a geometric progression ${{u}_{1}},{{u}_{2}},{{u}_{3}},{{u}_{4}},…$ are such that the sum to infinity is $81$ and the sum of the first $4$ terms is $80$.
If ${{u}_{1}}>100\,\,$and $\,\text{n}\ge \text{3}$,

(i)

Show that $\frac{3}{r}-\frac{6}{r+1}+\frac{3}{r+2}=\frac{6}{r\left( r+1 \right)\left( r+2 \right)}$.

[1]

(i) Show that $\frac{3}{r}-\frac{6}{r+1}+\frac{3}{r+2}=\frac{6}{r\left( r+1 \right)\left( r+2 \right)}$.

[1]

(ii)

Hence show that $\sum\limits_{r=1}^{N}{\frac{1}{r\left( r+1 \right)\left( r+2 \right)}}=\frac{1}{4}-\frac{1}{2\left( N+1 \right)}+\frac{1}{2\left( N+2 \right)}$.

[3]

(ii) Hence show that $\sum\limits_{r=1}^{N}{\frac{1}{r\left( r+1 \right)\left( r+2 \right)}}=\frac{1}{4}-\frac{1}{2\left( N+1 \right)}+\frac{1}{2\left( N+2 \right)}$.

[3]

(iii)

Give a reason why the series $\sum\limits_{r=1}^{\infty }{\frac{1}{r\left( r+1 \right)\left( r+2 \right)}}$ converges and write down its value.

[2]

(iii) Give a reason why the series $\sum\limits_{r=1}^{\infty }{\frac{1}{r\left( r+1 \right)\left( r+2 \right)}}$ converges and write down its value.

[2]

(iv)

Use your answer to (ii) to find $\sum\limits_{r=3}^{N}{\frac{1}{r\left( r-1 \right)\left( r-2 \right)}}$in terms of $N$.

[2]

(iv) Use your answer to (ii) to find $\sum\limits_{r=3}^{N}{\frac{1}{r\left( r-1 \right)\left( r-2 \right)}}$in terms of $N$.

[2]

##### 2019 DHS Promo Q2

Using the result $\sum\limits_{r=1}^{n}{\frac{r}{{{2}^{r}}}}=2-\frac{n+2}{{{2}^{n}}}$, show that $\sum\limits_{r=1}^{n}{\left( r-n \right)\left( {{2}^{-r}}+1 \right)}$ can be expressed in the form $C\left( 1-\frac{1}{{{2}^{n}}} \right)+Dn\left( n+1 \right)$, where $C$ and $D$ are constants to be determined.

[4]

##### 2017 VJC P1 Q8

It is given that $\sum\limits_{r=1}^{n}{\frac{{{r}^{2}}}{{{3}^{r}}}}=\frac{3}{2}-\frac{{{n}^{2}}+3n+3}{2\left( {{3}^{n}} \right)}$ .

(i)

Find $\sum\limits_{r=1}^{\infty }{\frac{{{r}^{2}}+{{\left( -1 \right)}^{r}}}{{{3}^{r}}}}$.

[3]

(i) Find $\sum\limits_{r=1}^{\infty }{\frac{{{r}^{2}}+{{\left( -1 \right)}^{r}}}{{{3}^{r}}}}$.

[3]

(ii)

Show that $\sum\limits_{r=4}^{n}{\frac{{{\left( r-2 \right)}^{2}}}{{{3}^{r-2}}}}=\frac{p}{q}-\frac{a{{n}^{2}}-an+a}{2\left( {{3}^{n-2}} \right)}$, where $a$, $p$ and $q$ are integers to be determined.

[5]

(ii) Show that $\sum\limits_{r=4}^{n}{\frac{{{\left( r-2 \right)}^{2}}}{{{3}^{r-2}}}}=\frac{p}{q}-\frac{a{{n}^{2}}-an+a}{2\left( {{3}^{n-2}} \right)}$, where $a$, $p$ and $q$ are integers to be determined.

[5]

##### 2020 EJC P1 Q10

The diagram shows the graph of $y=\frac{1}{{{x}^{2}}+1}$ when $x>0$.

(i)

Evaluate $\int_{k}^{k+1}{\frac{1}{{{x}^{2}}+1}\text{d}x}$ for $k>0$, leaving your answer in terms of $k$.

[2]

(i) Evaluate $\int_{k}^{k+1}{\frac{1}{{{x}^{2}}+1}\text{d}x}$ for $k>0$, leaving your answer in terms of $k$.

[2]

(ii)

By considering appropriate rectangles on the interval $\left[ k,k+1 \right]$for the curve $y=\frac{1}{{{x}^{2}}+1}$, show that

$\frac{1}{{{\left( k+1 \right)}^{2}}+1}<{{\tan }^{-1}}\left( k+1 \right)-{{\tan }^{-1}}k<\frac{1}{{{k}^{2}}+1}$ for $k\in {{\mathbb{Z}}^{+}}$.

[2]

(ii) By considering appropriate rectangles on the interval $\left[ k,k+1 \right]$for the curve $y=\frac{1}{{{x}^{2}}+1}$, show that

$\frac{1}{{{\left( k+1 \right)}^{2}}+1}<{{\tan }^{-1}}\left( k+1 \right)-{{\tan }^{-1}}k<\frac{1}{{{k}^{2}}+1}$ for $k\in {{\mathbb{Z}}^{+}}$.

[2]

(iii)

Use the identity $\tan \left( A-B \right)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ to show that

${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\frac{x-y}{1+xy}$, where $x>y>0$.

[2]

(iii) Use the identity $\tan \left( A-B \right)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ to show that

${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\frac{x-y}{1+xy}$, where $x>y>0$.

[2]

(iv)

By considering parts (ii) and (iii), prove by the method of differences that

$\sum\limits_{k=1}^{n}{\frac{1}{{{\left( k+1 \right)}^{2}}+1}}<{{\tan }^{-1}}\left( \frac{n}{n+2} \right)<\sum\limits_{k=1}^{n}{\frac{1}{{{k}^{2}}+1}}$

[4]

(iv) By considering parts (ii) and (iii), prove by the method of differences that

$\sum\limits_{k=1}^{n}{\frac{1}{{{\left( k+1 \right)}^{2}}+1}}<{{\tan }^{-1}}\left( \frac{n}{n+2} \right)<\sum\limits_{k=1}^{n}{\frac{1}{{{k}^{2}}+1}}$

[4]

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