Week -6 | Sequences & Series - 4

1.1 Introduction to infinite series’

We have introduced the concept of infinite series’ in our last modules. We have also briefly talked about the convergence of an infinite Geometric Series. The concept of convergence of a series is something that we have taken up in our advanced section of this module. To give an idea, here are some convergent infinite series expansions of popular functions:

$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+....$

$\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+....$

$e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....$

$a^x=1+(\ln a)x+(\ln a)^2 \frac{x^2}{2!}+....$

$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+...-1<x\le 1$

2.1 Combination of Series’

The series $S_n=a+(a+d)r+(a+2d)r^2+…$ is called an Arithmetico-Geometric Series. Note that the terms of this sequence are $a,(a+d)r,(a+2d)r^2,…$ , so this is clear that there are 2 sequences going on simultaneously, and the general term of this sequence is the product of the general terms of the 2 sequences.

We’ll show that the sum of this arithmetico-geometric series is given by

$S_n=\frac{a}{1-r}+dr\frac{1-r^{n-1}}{(1-r)^2}-\frac{\left \{ a+(n-1)d \right \}r^n}{1-r}$

We have assumed that $r \ne 1$ . In case where $r=1$ , the sequence becomes an AP.

We have :

$S_n=a+(a+d)r+(a+2d)r^2+...\left \{ a+(n-1)d \right \}r^{n-1}$

$\Rightarrow rS_n=ar+(a+d)r^2+(a+2d)r^3+...\left \{ a+(n-1)d \right \}r^n$

$\Rightarrow S_n(1-r)=a+(dr+dr^2+...dr^{n-1})-\left \{ a+(n-1)d \right \}r^n$

$\Rightarrow S_n(1-r)=a+dr\left( \frac{1-r^{n-1}}{1-r}\right )-\left \{ a+(n-1)d \right \}r^n$

The conclusion thus follows. This process of multiplying the sum by the common difference and then taking a difference is very useful in solving combination of series’ problems.

What happens when this sequence becomes infinite? As discussed before, for the convergence of an infinite GP series, $|r|<1$ . If we assume for now $nr^n\rightarrow 0$ as $n\rightarrow \infty$ (we’ll cover these proofs when we cover Limits in our Calculus module), then we can say that

$S_n\rightarrow \frac{a}{1-r}+\frac{dr}{(1-r)^2}$ as $n\rightarrow \infty$ .

2.2 Sigma Operator

Here are some useful properties of the sigma( $\sum$ ) operator:

$\sum\limits_{i=k}^n(T_i+T’_i)=\sum\limits_{i=k}^nT_i+\sum\limits_{i=k}^nT’_i$ if both the sums on the RHS exists.
$\sum\limits_{i=k}^n(T_iT’_i)\ne\left(\sum\limits_{i=k}^nT_i\right)\left(\sum\limits_{i=k}^nT’_i\right)$
$\sum\limits_{j=k}^n\sum\limits_{i=k}^n(T_iT_j)=\left(\sum\limits_{i=k}^nT_i\right)\left(\sum\limits_{j=k}^nT_j\right)$
$\sum\limits_{i=k}^naT_i=a\sum\limits_{i=k}^nT_i$

Ex-1

Given that $\sum\limits_{r=1}^nT_r=n(2n^2+9n+13)$ , find $\sum\limits_{r=1}^n\sqrt{T_r}$

Solution:

Let $S_n=\sum\limits_{r=1}^nT_r=n(2n^2+9n+13)$ , thus we have $T_r=S_r-S_{r-1}=6r^2+12r+6=6(r+1)^2$ . Thus $\sqrt{T_r}=\sqrt{6}(r+1)$ (we have taken the positive square root in this case).

$\sum\limits_{r=1}^n\sqrt{T_r}=\sqrt{6}\sum\limits_{r=1}^n(r+1)=\sqrt{6}\left(\sum\limits_{r=1}^nr+\sum\limits_{r=1}^n1 \right )$ . We thus have the sum as

$\sum\limits_{r=1}^n\sqrt{T_r}=\sqrt{6}\left(\frac{n^2+3n}{2} \right )=\sqrt{\frac{3}{2}}(n^2+3n)$ .

2.3 Method of Differences

Sometimes, in a series, we have the differences between the successive terms following a particular pattern. We find the sum of such series’ by the ‘method of differences’, Let’s consider an example:

Ex-2

Find the sum upto $n$ terms of this series: $3+15+35+63+…$

Solution:

Note that $15-3=12$ , $35-15=20$ , $63-35=28$ , so we have the difference between 2 successive terms in AP.

Thus let’s write this as

$S=3+15+35+63+..$ ,

$S=\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace \thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace\thinspace3+15+35+63+..$

Subtracting these 2 we get

$0=3+(12+20+28+…(n-1)$ terms $)-T_n$ . We thus have

$T_n=3+\frac{n-1}{2}[24+8(n-2)]=4n^2-1$ .

Thus $\sum\limits_{r=1}^nT_r=\sum\limits_{r=1}^n(4r^2-1)=\frac{n}{3}(4n^2+6n-1)$ .

Test Your Concepts:

Find the sum of the following series upto $n$ terms $5+7+13+31+85+…$

2.4 Method of Partial fractions

At times we may need to decompose the general term of a sequence into partial fractions to find the sum of the series. Consider the following series:

$S_n=\frac{1}{a(a+d)}+\frac{1}{(a+d)(a+2d)}+\frac{1}{(a+2d)(a+3d)}+...+\frac{1}{(a+(n-2)d)(a+(n-1)d)}$

We use the following decompositions to find the sum of this series:

$\frac{1}{a(a+d)}=\frac{1}{d}\left(\frac{1}{a}-\frac{1}{a+d} \right )$

$\frac{1}{(a+d)(a+2d)}=\frac{1}{d}\left(\frac{1}{a+d}-\frac{1}{a+2d} \right )$

$\frac{1}{(a+2d)(a+3d)}=\frac{1}{d}\left(\frac{1}{a+2d}-\frac{1}{a+3d} \right )$

and so on..

Thus the series becomes:

$S_n=\frac{1}{a(a+d)}+\frac{1}{(a+d)(a+2d)}+\frac{1}{(a+2d)(a+3d)}+...+\frac{1}{(a+(n-2)d)(a+(n-1)d)}$

$=\frac{1}{d}\left[\left(\frac{1}{a}-\frac{1}{a+d} \right ) +\left(\frac{1}{a+d}-\frac{1}{a+2d} \right )+...+\left(\frac{1}{a+(n-2)d} -\frac{1}{a+(n-1)d}\right )\right ]$

$=\frac{1}{d}\left[\frac{1}{a}-\frac{1}{a+(n-1)d} \right ]=\frac{n-1}{a(a+(n-1)d)}$

Ex -3

Find the Sum of the series upto $n$ terms:

$\frac{3}{1^2\times2^2}+\frac{5}{2^2\times3^2}+\frac{7}{3^2\times4^2}+...$

Solution:

From the problem statement it should be obvious that we are dealing with fractions – and the best way may be in this case will be the use of Partial fractions. So we start with the general term in this case:

We have $T_r=\frac{2r+1}{r^2(r+1)^2}=\left[\frac{1}{r^2}-\frac{1}{(r+1)^2} \right]$ .

Thus $\sum\limits_{r=1}^{n}T_r=\sum\limits_{r=1}^n\left[\frac{1}{r^2}-\frac{1}{(r+1)^2} \right ]= \left(\frac{1}{1^2}-\frac{1}{2^2} \right )+\left(\frac{1}{2^2}-\frac{1}{3^2} \right )+…+\left(\frac{1}{n^2}-\frac{1}{(n+1)^2} \right)$ .

This turns out to be $1-\frac{1}{(n+1)^2}$ .

Test Your Concepts:

Find the sum of the following series’ upto $n$ terms:

$\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+…$ . This may be a good time to learn a very useful identity – Sophie Germain Identity.Can you try to factorise $r^4+r^2+1$ in a similar way?
$1+\frac{1}{1+2}+\frac{1}{1+2+3}+…\frac{1}{1+2+3+..+n}$

Week -6 | Sequences & Series – 4

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