### <Success>Infinite series lecture 1</Success> <div style='float: left;text-align:left; width:100%;'> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-13-infinite-series-l-1_5-k98edzv0vh0/img/014-0027.6.jpg"> <img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/b2eb2eac-8f06-4e7e-ac47-50acf235db00/public"></p> </div>

### <Success>Weierstrass approximation theorem</Success> <div style='float: left;text-align:left; width:70%;font-size:30px;'> <p>If f is a continous real-valued function defined on the closed interval [a,b], then for every $x$ $\in$ [a,b]<br> and given any $\epsilon > 0$, there exists a polynomial</p> <p>$p $ $ \ \exists \forall x \epsilon$ [a,b]</p> <p>$|f(x) - p(x)| < \epsilon $</p> <p>$p_n(x) = a_0 + a_1x + …….+ a_n x^n$</p> </div> <div style='float: right; width:30%;'> <img src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/431448cf-b49a-4a9a-ba06-ad6b0fa3c500/public"> <!----> </div>

### <Success>Weierstrass approximation theorem</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>In other words,</p> <p>We can approximate $f(x)$ by a polynomial $p(x)$ .</p> <p>consider $\sin (x)$ . If we can find a polynomial $f^n$ to approximate $\sin x$, then we are done.</p> <p>The question is how to obtain such a polynomial?</p> <p>We have to look at a series.</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/b5fc21bc-9d7c-4086-6d32-ee83a2228400/public"></p> </div>

### <Success>Sequence</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>An ordered arrangements of certain objects.</p> <p>Example: Alphabet: A,B,C……$\Z$</p> <p>These 26 letters come in that specific order.</p> <p>Rainbow colours: V I B G Y O R</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/0179a649-6031-44f2-08e7-d82fb7229800/public"></p> </div>

### <Success>Example</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Prime numbers:</p> <p>2,3,5,7,11,13…………</p> <p>Even numbers:</p> <p>2,4,6,8,10,……………….</p> <p>Two things we have to observe.</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-13-infinite-series-l-1_5-k98edzv0vh0/img/137-0751.2.jpg"></p> </div>

### <Success>Sequence</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Sequence may be finite or infinite.</p> <ul> <li>Alphabet, colours of Rainbow etc are finite sequence.</li> <li>i.e. the number of terms is finite. All odd numbers between 0 to 100.</li> </ul> <p>Viz $ 1,3,5,………………99$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/38c414d2-73ec-436f-d540-d722020ef700/public"></p> </div>

### <Success>Infinte sequence</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>The number of terms can be arbitrarily large.</p> <p>i.e. Choose any large number $N$,<br> The number of terms in the sequence in $ > N$.</p> <p><strong>Example:</strong></p> <ul> <li>All even numbers.</li> <li>All powers of 2.<br> $2,4,8,16,32,………..$</li> </ul> <p>i.e. given any $k$, $2^k$ belongs to the above sequence.</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/8ec9f7a5-2bd8-49b1-6a63-d97a7092e500/public"></p> </div>

### <Success>Limit of Sequence</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>The second issue is whether the sequence has a limit.</p> <p><strong>Example:</strong></p> <ol> <li> <p>1,2,3,……. sequence of natural numbers. Does it have a limit?</p> </li> <li> <p>$2^k : k = 1,2,………..$<br> Does not have any limit.</p> </li> </ol> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-13-infinite-series-l-1_5-k98edzv0vh0/img/217-1095.6.jpg"></p> </div>

### <Success>Example 3</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$ 1 + \frac {1}{k}$, $k = 1,2,3,……..$</p> <p>$k = 1$, the value is $2$</p> <p>$k = 2$, the value is $1 + \frac {1}{2} = 1.5$</p> <p>$\vdots$</p> <p>$k = 10 = 1.1$</p> <p>$k = 100 = 1.01$</p> <p>$k = 1000 = 1.001$</p> <p>It is always $> 1$, but as $k$ increases, it is close to $1.$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-13-infinite-series-l-1_5-k98edzv0vh0/img/248-1260.0.jpg"></p> </div>

### <Success>Example 4</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Consider $(-1)^n$</p> <p>$(-1)^1 , (-1)^2, (-1)^3, (-1)^4,…………..$</p> <p>$= -1, +1, -1, +1,………………$</p> <p>$\therefore$ As $n$ increases the values of the sequence are not converging to any value $r.$</p> <p>$\therefore$ This sequence also does not have any limit.</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/69956d4e-b46c-479e-31ab-7a409ca17000/public"></p> </div>

### <Success>Series</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Given a sequence</p> <p>$a_1, a_2, a_3$,………….</p> <p>Where $a_k$ is the $k^{th}$ term of the sequence.</p> <p>The Summation $a_1 + a_2 + a_3 +……………$</p> <p>or $\underset {k=1,2,3…..}\sum a_k$ is called the series.</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/41271d76-b220-437e-0946-2a147abecf00/public"></p> </div>

### <Success>Series</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>If the sequence is finite and the $a_k$ are numeric then we can obtain the value of the series or sum of the series.</p> <p>In fact you have seen A.P, G.P, etc and we have obtained the sum of the first $n$ terms.</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/ccc81de5-8238-4067-5fa7-15c11b1ddf00/public"></p> </div>

### <Success>Question</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>What happens if the sequence is infinite?</p> <p>Example: $a_k = 2^k$</p> <p>$a_k = 5 - 2k$</p> <p>$a_k = \sin (2k \pi)$</p> <p>$a_k = \sin \frac {1}{2k \frac{\pi}{2}}$</p> <p>The series will be $\underset {k=1} \sum^{\infty} a_k.$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-13-infinite-series-l-1_5-k98edzv0vh0/img/000001.png"></p> </div>

### <Success>Question</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>If the series is infinite does it converge to some value?</p> <p><strong>Solution:</strong></p> <p>we try to answer it the following way:</p> <p>Let $S_n = \underset {k=1} \sum^{n} a_k$</p> <p>This is called the $n^{\text{th}}$ partial sum.</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-13-infinite-series-l-1_5-k98edzv0vh0/img/000002.png"></p> </div>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$S_1 = a_1$</p> <p>$S_2 = a_1 + a_2$</p> <p>$S_3 = a_1 + a_2 + a_3$</p> <p>$S_n = a_1 + a_2 + …+ a_n$</p> <p>Thus, we generate a sequence $S_1, S_2, ……., S_n$</p> <p>This series will be convergent if the corresponding sequence of partial sums $S_1, S_2, ……., S_n$ has a limit.</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-13-infinite-series-l-1_5-k98edzv0vh0/img/365-1928.4.jpg"></p> </div>

### <Success>Example</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$G_k = k$</p> <p>Then $S_n = 1 + 2 + ….+ n = \frac {n(n+1)}{2}$</p> <p>As $n\longrightarrow \infty$ , The $S_n \longrightarrow \infty$</p> <p>$\therefore $ The series $\underset {k=1} \sum ^ {\infty}$ k does not converge.</p> </div> <div style='float: right; width:0%;'> </div>

### <Success>Example</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$G_k = (-1)k$</p> <p>$a_1 = -1 , \quad S_1 = -1$</p> <p>$a_2 = +1 , \quad S_1 = 0$</p> <p>$a_3 = -1, \quad S_3 = -1$</p> <p>$a_4 = +1, \quad S_4 = 0$</p> <p>This series does not converge.</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/fdd3a335-9b5c-429e-38cb-4dd5e161cb00/public"></p> </div>

### <Success>Convergence of the series</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Let us consider G.P</p> <p>$a, ar, ar^2, ar^3,…………$</p> <p>i.e. the $k^{th}$ term $= a r^{k-1}$</p> <p>$S_k = a + ar + ar^2 + …….+ ar^{k-1}$</p> <p>$\therefore r S_k = - ar + - ar^2 + …..+ a r^{k-1} + a r^{k}$</p> <p>i.e. $S_k - r S_k = a - a r^k = a(1-r^k)$<br> $(1-r)S_k = a(1-r^k)$</p> <p>$S_k = \frac {a(1-r^k)}{1-r}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-13-infinite-series-l-1_5-k98edzv0vh0/img/000003.png"></p> </div>

### <Success>Convergence of the series</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>We find that for the G.P</p> <p>$ S_k = \frac {a(1-r^k)}{1-r} , or, \frac {a (r^k-1)}{r-1}$</p> <p>What happens if k$\longrightarrow \infty$</p> <p>Consider $\underset {k\longrightarrow \infty}\lim \frac {a(1-r^k)}{1-r}$</p> <p>if |r| < 1 then we know $r^k \longrightarrow$ 0</p> <p>$\frac {a(1-r^k)}{1-r} \longrightarrow \frac {a}{1-r} as, k \longrightarrow \infty$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-13-infinite-series-l-1_5-k98edzv0vh0/img/459-2412.0.jpg"></p> </div>

### <Success>Convergence of the series</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>On the other hand if $|r| > 1$</p> <p>then $r^k \longrightarrow \infty$</p> <p>$\therefore \underset {k \rightarrow \infty}\lim \frac {a(r^k -1)}{r-1} \rightarrow \infty$</p> <p>The series does not converge.</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-13-infinite-series-l-1_5-k98edzv0vh0/img/000004.png"></p> </div>

### <Success>Convergence of the series</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>If we look at $a, ar, ar^2 + …………$</p> <p>$\equiv a(1 + r + r^2 + ……..)$</p> <p>It will converge if $|r| < 1$ else it will diverge.</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-13-infinite-series-l-1_5-k98edzv0vh0/img/000005.png"></p> </div>

### <Success>Convergence of the series</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$1 + r + r^2 + ………….$</p> <p>where does it converge $\frac {a(1-r^k)}{1-r}$ G.P series upto $k$ term.</p> <p>If $|r| < 1$ the series sum will be $\frac {a}{1-r}$</p> <p>In particular , keeping $a = 1 $</p> <p>we can say:</p> <p>$1 + r + r^2 + r^3 + …….= \frac {1}{1-r}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-13-infinite-series-l-1_5-k98edzv0vh0/img/000006.png"></p> </div>

### <Success>Example</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <ul> <li>$1 + \frac{1}{2} + \frac {1}{4} + \frac {1}{4} + \frac {1}{8} + ……. = \frac {1}{1-\frac{1}{2}} = 2$</li> <li>$1 + \frac {1}{3} + \frac{1}{9} + \frac{1}{27}+……. \underset {k=0}\sum ^{\infty} \frac {1}{3k} = \frac {1}{1-\frac{1}{3}} = \frac {3}{2}$</li> </ul> <p>$\therefore$ We can say :</p> <p>$(1-r)^{-1} = 1 + r + r^2 + ……….$</p> <p>for $|r| < 1$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/a5232ec4-803b-4c6a-3816-a5ab7d597500/public"></p> </div>

### <Success>Question</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>What is $(1+r)^{-1}?$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-13-infinite-series-l-1_5-k98edzv0vh0/img/550-2815.2.jpg"></p> </div>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>We are looking at $(1+r)^{-1}$</p> <p>We can write it as : $(1-(-r))^{-1}$</p> <p>$\therefore$ Putting $r = -r$ in the series expansion.</p> <p>$(1+r)^{-1} = 1 + (-r) + (-r)^2 + (-r)^3 + ………$</p> <p>$ = 1 - r + r^2 - r^3 + r^4……….$</p> <p>$|r| < 1$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-13-infinite-series-l-1_5-k98edzv0vh0/img/000007.png"></p> </div>

### <Success>Recall</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Have you seen similar expression before?</p> <p><strong>Binomial expansion:</strong> What is $(1+r)^n \leftarrow$</p> <p>$= 1 +{ }^nC_1 r +{ }^nC_2 r^2 + …….+ { }^nC_n r^n$</p> <p>Where ${ }^nC_k$ = $\frac {n!}{k!(n-k)!}$</p> <p>Here $n$ is a positive integer.</p> <ul> <li>We know how to expand $(1+r)^n$ for a given positive integer n.</li> <li>We know that there are n + 1 terms in the series since it is finite we can compute its value.</li> </ul> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-13-infinite-series-l-1_5-k98edzv0vh0/img/000008.png"></p> </div>

### <Success>Summary</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Today we have seen</p> <ul> <li>$(1+r)^{-1} \leftarrow $ the power is negative. <ul> <li>for $(+)$ve r $\rightarrow$ finite series.</li> <li>for $(-)$ve r $\rightarrow $ we get an infinite series.</li> </ul> </li> <li>$(1-r)^{-1} or (1+r){-1}$</li> </ul> <p>we could compute their value if $|r| < 1.$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/86efe3a1-9054-4d80-547e-cb55d1684a00/public"></p> </div>

### <Success>Thank you</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> </div> <div style='float: right; width:0%;'>