### <Success>Infinite series lecture 5</Success> <div style='float: 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-17-infinite-series-l-5_5-vyi-pog9ubk/img/014-0027.6.jpg"> <img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/8eee2844-5eaa-4c79-5890-c5a766deae00/public"></p> </div>

### <Success>Recall</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Euler’s Number</p> <p>$ e=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^n $</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/019-0100.8.jpg"></p> </div>

### <Success>Recall</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Consider $\left(1+\frac{1}{n}\right)^n$</p> <p>The $k^{\text {th }}$ term for $k \leqslant n$ can be obtained from Binomial Theorem.</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/025-0152.4.jpg"></p> </div>

### <Success>Recall</Success> <div style='float: left; width:100%;font-size:30px;'> <p>The $k^{\text {th }}$ term for $k=$ $0,1,2, \ldots n$ is</p> <p>$ {}^n C_k \left( \frac{1}{n} \right)^k$</p> <p>$= \frac{n!}{k !(n-k) !} \frac{1}{n^k} $</p> <p>$= \frac{n(n-1) \cdots(n-\overline{k-1)}}{k !} \frac{1}{n^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-17-infinite-series-l-5_5-vyi-pog9ubk/img/036-0230.4.jpg"></p> </div>

### <Success>Recall</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$ =\frac{1\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) \cdots\left(1-\frac{k-1}{n}\right)}{k !} $</p> <p>$ \text { As } n \rightarrow \infty $ $\frac{1}{k!}$</p> <p>$ \therefore \text { The limit of }\left(1+\frac{1}{n}\right)^n $</p> <p>$ \rightarrow \frac{1}{0 !}+\frac{1}{1 !}+\cdots+\frac{1}{k !}+ \cdots $</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/000001.png"></p> </div>

### <Success>Recall</Success> <div style='float: left; width:100%;font-size:30px;'> <p>The infinite series can be seen as</p> <p>$ 1+1+\frac{1}{2 !}+\frac{1}{3 !}+\cdots \cdot \cdot $</p> <p>This is called Euler’s Number.</p> <p>This is an irrational number</p> <p>$2 \cdot {71828}1828459045 230365$</p> <p>$2 < e$</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/000002.png"></p> </div>

### <Success>Question 1</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Check if $e$ is greater than 3 or not.</p> <p><strong>Solution:</strong></p> <p>Consider $\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+ \cdots$</p> <p>$=\frac{1}{2}+\frac{1}{2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{2 \cdot 3 \cdot 4 \cdot 5} +\cdots$</p> <p>$=\frac{1}{2}+\frac{1}{2 \cdot 2}+\frac{1}{2 \cdot 2 \cdot 2}+\frac{1}{2^4} +\cdots $</p> <p>$=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots $</p> <p>$=\frac{1}{2}\left(\frac{1}{1-\frac{1}{2}}\right)=1$</p> <p>$2 < e < 2+1=3$</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/133-0735.6.jpg"></p> </div>

### <Success>$\text{Expansion of} \ e^x$</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Let us consider $e^x$ for some real or complex $x.$ What is the series for $e^x$ ?</p> <p>$ e^x=1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+ $</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/155-0816.0.jpg"></p> </div>

### <Success>$\text{Expansion of} \ e^x$</Success> <div style='float: left; width:100%;font-size:30px;'> <p><strong>Intuitive idea</strong><br> Consider $e^2 = e \times e$</p> <p>$ \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{2 n} $</p> <p>$= \lim _{n \rightarrow \infty}\left(\left(1+\frac{1}{n}\right)^2\right)^n $</p> <p>$= \lim _{n \rightarrow \infty}\left(1+\frac{2}{n}+\frac{1}{n^2}\right)^n$</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/170-0902.4.jpg"></p> </div>

### <Success>$\text{Expansion of} \ e^x$</Success> <div style='float: left; width:100%;font-size:30px;'> <p>The $k^{\text {th }}$ term for $0 \leq k \leq n$</p> <p>is $ {}^n C_k\left(\frac{2}{n}+\frac{1}{n^2}\right)^k $</p> <p>$= \frac{n !}{k !(n-k) !} \frac{1}{n^k}\left(2+\frac{1}{n}\right)^k$</p> <p>$= \frac{1\left(1-\frac{1}{n}\right) \cdots\left(1-\frac{k-1}{n}\right)}{k !}\left(2+\frac{1}{n}\right)^k$</p> <p>$\lim _{n \rightarrow \infty} \frac{1}{k !} 2^k=\frac{2^k}{k !}$<br> (for a fixed $k$ )</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/196-1029.599.jpg"></p> </div>

### <Success>$\text{Expansion of} \ e^x$</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\therefore e^2= \frac{2^0}{0 !}+\frac{2^1}{1 !}+\cdots+\frac{2^k}{k !}+\cdots $</p> <p>$ \text { or } 1+\frac{2}{1 !}+\frac{2^2}{2 !}+\cdots+\frac{2^k}{k !}+\cdots $</p> <p>This gives the idea for</p> <p>$ e^x=1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+ \cdots $</p> <p>$ e^{-x}=1-x+\frac{x^2}{2 !}-\frac{x^3}{3 !} +\cdots $</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/e0aaf3a4-42da-4210-8b4d-fbdfc7bd3900/public"></p> </div>

### <Success>Remark</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Considers the power $e^{i x}$</p> <p>$=1+i x+\frac{(i x)^2}{2 !}+\frac{(i x)^3}{3 !}+\frac{(i x)^4}{4 !} +\frac{(i x)^5}{5 !}+\frac{(i x)^6}{6 !}+\cdots $</p> <p>$ =1+i x-\frac{x^2}{2 !}-i \frac{x^3}{3 !}+\frac{x^4}{4 !}+i \frac{x^5}{5 !} \cdots$</p> <p>$ = \underbrace {\left(1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\frac{x^6}{6 !} \cdots \right)}_ {=\cos x} + i \underbrace {\left(x-\frac{x^3}{3 !}+\frac{x^5}{5 !} \cdots \right)}_{=\sin x}$</p> <p>$e^{i x} = \cos x+i \sin x$</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/256-1374.0.jpg"></p> </div>

### <Success>Question 2</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Find the value of</p> <p>$ \frac{1}{1 !}+\frac{2}{2 !}+\frac{3}{3 !}+\cdots . $</p> <p><strong>Solution:</strong></p> <p>$\frac{1}{1 !}+\frac{2}{2 !}+\frac{3}{3 !}+\cdots = \frac{1}{1 !}+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !} \cdots .$</p> <p>This is also $=e$.</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/275-1460.4.jpg"></p> </div>

### <Success>Example</Success> <div style='float: left; width:100%;font-size:30px;'> <p><strong>1)</strong> $\frac{x}{1 !}+\frac{2 x}{2 !}+\frac{3 x}{3 !}+ \cdots$</p> <p>$ =x\left(\frac{1}{1 !}+\frac{2}{2 !}+\frac{3}{3 !}+\cdots\right)=x e $</p> <p><strong>2)</strong> $\sum_{n=0}^{\infty} \frac{1}{(n-1) !}=\sum_{n=1}^{\infty} \frac{1}{(n-1) !}$</p> <p>$=\sum_{m=0}^{\infty} \frac{1}{m !}$ where $m=x-1$</p> <p>$=e$.</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/000003.png"></p> </div>

### <Success>Example 3</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$ \sum_{n=0}^{\infty} \frac{1}{(n-2) !}=\sum_{n=2}^{\infty} \frac{1}{(n-2) !}$</p> <p>$=\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\cdots $</p> <p>$ =e$</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/308-1658.4.jpg"></p> </div>

### <Success>Example 4</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\sum_{i=0}^{\infty} \frac{i^2}{i !}=\sum_{i=1}^{\infty} \frac{i^2}{i !}$</p> <p>The $k^{\text {th }}$ term $=\frac{k^2}{k !}=\frac{k}{(k-1) !}$</p> <p>$=\left(\frac{k-1+1}{(k-1) !}\right)=\frac{k-1}{(k-1) !}+\frac{1}{(k-1) !} $</p> <p>$=\sum_{k=2}^{\infty} \frac{1}{(k-2) !}+\sum_{1}^{\infty} \frac{1}{(k-1) !}$</p> <p>$ =e+e=2 e$</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/329-1812.0.jpg"></p> </div>

### <Success>Example 5</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\sum_{n=0}^{\infty} \frac{n^3}{n !}=\sum_{n=1}^{\infty} \frac{n^2}{(n-1) !} $</p> <p>$=\sum_{n=1}^{\infty} \frac{(n-1)^2+2 n-1}{(n-1) !} $</p> <p>$=\sum_{n=1}^{\infty} \frac{(n-1)}{(n-2) !}+2 \sum^{\infty} \frac{n}{(n-1) !}-\sum_{n=1}^{\infty} \frac{1}{(n-1)} $</p> <p>$=\sum_n \frac{(n-2)+1}{(n-2) !}+2 \sum \frac{n-1+1}{(n-1) !}-\sum \frac{1}{(n-1)}$</p> <p>$ =\sum \frac{1}{(n-3) !}+\sum \frac{1}{(n-2) !}+2 \sum \frac{1}{(n-2) !} $</p> <p>$ +2 \sum \frac{1}{(n-1) !}-\frac{1}{(n-1) !}$</p> <p>$= e + e + 2e + 2e -e = 5e $</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/349-2119.2.jpg"></p> </div>

### <Success>Example 6</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Consider $\frac{1}{1 !}+\frac{1+2}{2 !}+\frac{1+2+3}{3 !}+\cdots$</p> <p>We can see the $k^{\text {th }}$ term is $\frac{\sum_{1}^{k} i}{k !}=\frac{\frac{k(k+1)}{2}}{k !}$</p> <p>$= \frac{1}{2}\left(\frac{k}{k !}+\frac{k+1}{k !}\right)$</p> <p>$=\frac{1}{2}(\frac{1}{(k-1) !}+ \frac{1}{(k-1) !}+\frac{1}{k !})$</p> <p>$\therefore$ Taking sum, $\sum_{i=1}^{\infty} \frac{1+2+3+\cdots=i}{i!}$</p> <p>$ =\frac{1}{2}\left(\sum \frac{1}{(k-1) !}+\sum \frac{1}{(k-1) !}+\sum \frac{1}{k !}\right) $<br> $=\frac{3}{2} e$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/7d365c7a-5cee-48a1-39e9-3d7e5a645800/public"></p> </div>

### <Success>Question 3</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Find the coefficient of $x^4$ in $\left(1+2 x+3 x^2\right) e^{-x}$.</p> <p><strong>Solution:</strong></p> <p>$ \left(1+2 x+3 x^2\right)(1-x+\frac{x^2}{2 !}-\frac{x^3}{3 !}+\frac{x^4}{4 !}\cdots) $</p> <p>$ =\frac{1}{4 !}-\frac{2}{3 !}+\frac{3}{2 !} $</p> <p>$ =\frac{1}{24}-\frac{2}{6}+\frac{3}{2} $</p> <p>$ =\frac{1-8+36}{24}=\frac{29}{24}$</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/000004.png"></p> </div>

### <Success>Question 4</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Find the value of</p> <p>$ 1+5 ln 3+\frac{(5 ln 3)^2}{2 !}+ \cdots $</p> <p>$= e^{5 ln 3}$</p> <p>$= e^{ln 3^5}$</p> <p>$= 3^5$</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/429-2576.4.jpg"></p> </div>

### <Success>Question 6</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Find the expansion for $\frac{e^x}{\cos x}.$</p> <p><strong>Solution:</strong></p> <p>Let the corresponding series be</p> <p>$ C_0+C_1 x+C_2 x^2+C_3 x^3+ … $<br> $e^x=\cos x(C_0+C_1 x +C_2 x^2+C_3 x^3 +\ldots)$ $\left(1+x+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\cdots\right) $</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/449-2685.6.jpg"></p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$ =\left(1-\frac{x^2}{2 !}+\frac{x^4}{4 !}\cdots \right)$</p> <p>$\left(C_0+C_1 x + +C_2 x^2+ C_3 x^3+\cdots\right) $</p> <table> <thead> <tr> <th>$x^0$</th> <th>1</th> <th>$C_0\times1$</th> <th>$\Rightarrow C_0=1$</th> </tr> </thead> <tbody> <tr> <td>$x^1$</td> <td>1</td> <td>$C_1 \times 1$</td> <td>$\Rightarrow C_1=1$</td> </tr> <tr> <td>$x^2$</td> <td>$\frac{1}{2}$</td> <td>$C_2\times1- \frac{C_0}{2}$</td> <td>$\Rightarrow C_2-\frac{1}{2}=\frac{1}{2}$</td> </tr> <tr> <td>$x^3$</td> <td>$\frac{1}{3!}$</td> <td>$C_3- \frac{C_1}{2}$</td> <td>$\Rightarrow \frac{1}{6}=C_3-\frac{1}{2}$</td> </tr> </tbody> </table> <p>$\therefore C_2 =1$</p> <p>$\therefore C_3 =\frac{2}{3}$</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/000005.png"></p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>The coefficient & $\frac{e^x}{\cos x}$ are</p> <p>$ C_0=1 $</p> <p>$ C_1=1 $</p> <p>$ C_2=1 $</p> <p>$ C_3=\frac{2}{3}$</p> <p>$ C_4=\frac{1}{2} $</p> <p>$ C_5=\frac{3}{10}$<br> $\vdots$</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-17-infinite-series-l-5_5-vyi-pog9ubk/img/510-3004.8.jpg"></p> </div>

### <Success>Thank you</Success> <div style='float: left; width:100%;font-size:30px;'> </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-17-infinite-series-l-5_5-vyi-pog9ubk/img/513-3040.8.jpg"></p>