### <Success>Infinite Series lecture 4</Success> <div style='float: left; text-align: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-16-infinite-series-l-4_5-hviteb7nlgk/img/016-0044.4.jpg"></p> </div>

### <Success>Recall</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p><strong>Taylor series expansion</strong></p> <p>$ f(x)=f(a) +f^{(1)}(a) \frac{(x-a)}{1 !}$</p> <p>$ \ldots +f^{(2)}(a) \frac{(x-a)^2}{2 !}$</p> <p>$ \ldots +f^{(k)}(a) \frac{(x-a)^k}{k !}$</p> <!--- $\sin x$ --> <!--- $\cos x$--> <!--- $\tan x$--> </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-16-infinite-series-l-4_5-hviteb7nlgk/img/030-0134.4.jpg"></p> </div>

### <Success>Question 1</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Find the Taylor Series approximation of $\tan ^{-1}(x)$ up to $x^5$.</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-16-infinite-series-l-4_5-hviteb7nlgk/img/050-0237.6.jpg"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$ f(x) =\tan ^{-1}(x) \quad f(0)=0$</p> <p>$f^{\prime}(x) =\frac{1}{1+x^2}=\left(1+x^2\right)^{-1} \quad \therefore f^{\prime} (0)=1$</p> <p>$ f^{(2)}(x) =\frac{d}{d x}\left(1+x^2\right)^{-1}$</p> <p>$ =-\left(1+x^2\right)^{-2}(2 x)$</p> <p>$=-2 x\left(1+x^2\right)^{-2}$</p> <p>$ \therefore f^{(2)}(x)=0 \text { at } x=0$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/906c12b5-684a-4bf5-e6ae-e35769e4ae00/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$ f^{(3)}(x)=\frac{d}{d x}\left(-2 x\left(1+x^2\right)^{-2}\right)$</p> <p>$=-2\left(1+x^2\right)^{-2}+(-2 x)(-2)(1+x^2)^{-3}(2 x)$</p> <p>$=-2\left(1+x^2\right)^{-2}+8 x^2\left(1+x^2\right)^{-3}$</p> <p>$\therefore f^{(3)}(0)=-2$</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-16-infinite-series-l-4_5-hviteb7nlgk/img/000001.png"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$ f^{(4)}(x)=\frac{d}{d x}(-2(1+x^2)^{-2}+8 x(1+x^2)^{-3})$</p> <p>$=4(1+x^2)^{-3}(2 x)+16 x (1+x^2)^{-3} + 8 x^2 + (1+x^2 )^{-4} 2 x$</p> <p>$=8 x\left(1+x^2\right)^{-3}+16 x\left(1+x^2\right)^{-3} +8 x^3\left(1+x^2\right)^{-4} 2x$</p> <p>$=24 x\left(1+x^2\right)^{-3}+8 x^3\left(1+x^2\right)^{-4}$</p> <p>$\therefore \quad f^{(4)}(0)=0$</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-16-infinite-series-l-4_5-hviteb7nlgk/img/093-0618.0.jpg"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$ f^{(5)}(x)=\frac{d}{d x} \left(24 x\left(1+x^2\right)^{-3} +8 x^3\left(1+x^2\right)^{-4}\right)$</p> <p>$=24\left(1+x^2\right)^{-3}+24 x\left(1+x^2\right)^{-4}(2 x)+ \ldots $</p> <p>$ f^{(5)}(0)=24$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/14d31720-bf3e-40b2-fefb-b724a5f09200/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$f(0)=0$<br> $f^{(1)}(0)=1$<br> $f^{(2)}(0)=0$<br> $f^{(3)}(0)=-2$<br> $f^{(4)}(0)=0$<br> $f^{(5)}(0)=24$</p> <p>$\therefore$ The $5^{\text {th }}$ degree polynomial approximation for $\tan ^{-1}(x)$ is:</p> <p>$ x-\frac{2 x^3}{3 !}+\frac{24 x^5}{5 !}.$</p> <p>$ \equiv x-\frac{x^3}{3}+\frac{x^5}{5} .$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/fc42d3cd-9415-40ea-5f63-329491ba8c00/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$ \frac{d}{d x} \tan ^{-1}(x)=\frac{1}{1+x^2} $</p> <p>The expansion of $\frac{1}{1+x^2}$</p> <p>$\text { i.e }\left(1+x^2\right)^{-1}$</p> <p>$1-x^2+x^4-x^6 \ldots$</p> <p>$ \int \frac{1}{1+x^2} d x=\int 1 d x-\int x^2 d x + \int x^4 dx \ldots +C$</p> <p>$ \tan ^{-1} x=x-\frac{x^3}{3}+\frac{x^5}{5} \ldots+C$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/4a2fe2e4-3094-4975-f718-1e396f90e800/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>By putting $x=0$</p> <p>we find that $c=0$</p> <p>$ x-\frac{x^3}{3}+\frac{x^5}{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-16-infinite-series-l-4_5-hviteb7nlgk/img/151-1044.0.jpg"></p> </div>

### <Success>$\text{Expansion of} \ \sin ^{-1}x$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Consider $\sin ^{-1} x$</p> <p>we knows that $\frac{d}{d x} \sin ^{-1}(x)$</p> <p>$ =\frac{1}{\sqrt{1-x^2}}$</p> <p>$ =\left(1-x^2\right)^{-\frac{1}{2}}$</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-16-infinite-series-l-4_5-hviteb7nlgk/img/170-1120.8.jpg"></p> </div>

### <Success>$\text{Expansion of} \ \sin ^{-1}x$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$ \left(1-x^2\right)^{-\frac{1}{2}}$</p> <p>$=1 +\left(-\frac{1}{2}\right)\left(-x^2\right)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)}{2 !}\left(-x^2\right)^2$</p> <p>$ +\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)}{3 !}\left(-x^2\right)^3$</p> <p>$= 1+\frac{x^2}{2}+\frac{1 \cdot 3}{8} x^4+\frac{1 \cdot 3 \cdot 5}{8 \times 3 !} x^6$</p> <p>$= 1+\frac{x^2}{2}+\frac{3}{8} x^4+\frac{15^4}{48} x^6+\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-16-infinite-series-l-4_5-hviteb7nlgk/img/188-1260.0.jpg"></p> </div>

### <Success>$\text{Expansion of} \ \sin ^{-1}x$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$ \therefore \int \frac{1}{\sqrt{1-x^2}} d x$</p> <p>$ \quad=\int 1 d x+\int \frac{x^2}{2} d x+\int \frac{3}{8} x^4 d x +\int \frac{15}{48} x^6 d x \ldots +c$</p> <p>$ \text { or } \sin ^{-1}(x)=x+\frac{x^3}{3 !}+\frac{3 x^5}{40}+\frac{15}{48 \times 7} x^7 \ldots +c$</p> <p>Putting $x=0$</p> <p>we get $c=0$</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-16-infinite-series-l-4_5-hviteb7nlgk/img/203-1364.4.jpg"></p> </div>

### <Success>$\text{Expansion of} \ \sin ^{-1}x$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\therefore$ Taylor series expansion for $\sin ^{-1}(x)$</p> <p>$=x^4+\frac{x^3}{3 !}+\frac{3 x^5}{40}+\frac{5}{16 \times 7} x^7 \cdots$</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-16-infinite-series-l-4_5-hviteb7nlgk/img/211-1420.8.jpg"></p> </div>

### <Success>$\text{Expansion of} \ \cos ^{-1}x$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>What is the expansion for $\cos ^{-1}(x)$?</p> <p>$ f^{\prime} f^{\prime \prime} f^{\prime \prime \prime} \ldots $</p> <p>$ \sin ^{-1} x=\frac{\pi}{2}-\cos ^{-1} x$</p> <p>$ \text { or } \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} 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-16-infinite-series-l-4_5-hviteb7nlgk/img/228-1485.6.jpg"></p> </div>

### <Success>$\text{Expansion of} \ \cos ^{-1}x$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$ \therefore \cos ^{-1}(x)$</p> <p>$ =\frac{\pi}{2}-x-\frac{x^3}{3 !}-\frac{3 x^5}{40} -\frac{5 x^7}{16 \times 7} \ldots $</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-16-infinite-series-l-4_5-hviteb7nlgk/img/234-1526.4.jpg"></p> </div>

### <Success>$\text{Expansion of} \ log (1+x)$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Let us consider ${log (1+x)}$</p> <p>$ f(x)=\log (1+x)\quad \therefore f(0)=\log (1)=0$</p> <p>$ f^{\prime}(x)=\frac{1}{1+x} \quad\therefore f^{\prime}(0)=1$</p> <p>$ f^{\prime \prime}(x)=-(1+x)^{-2}\quad \therefore f^{\prime \prime}(0)=-1$</p> <p>$ f^{\prime\prime\prime}(x)=2(1+x)^{-3}\quad \therefore f^{\prime\prime\prime}(0)=2 $</p> <p>$ f^{\prime\prime\prime\prime}(x)=-6(1+x)^{-4} \quad\therefore f^{\prime\prime\prime\prime}(0)=-6$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/dfe2fbe1-3a95-4f15-15bb-eb2ef9403200/public"></p> </div>

### <Success>$\text{Expansion of} \ log (1+x)$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\therefore \log (1+x)$</p> <p>$ x-\frac{x^2}{2}+2 \frac{x^3}{3 !}-6 \frac{x^4}{4 !} \cdots$</p> <p>$= x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4} \cdots$</p> <p>$\therefore$ The series is: $\sum_{k=1}^{\infty} \frac{x^k}{k}(-1)^{k-1}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/04f22548-69f8-4e82-cfb0-743a5f2a9200/public"></p> </div>

### <Success>$\text{Expansion of} \ log (1-x)$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>Since we know expansion of $\frac{1}{1+x}$</p> <p>$\frac{1}{1+x} =(1+x)^{-1}$</p> <p>$ =1-x+x^2-x^3+x^4 \ldots$</p> <p>$\int \frac{1}{1+x} d x =\int 1 d x-\int x d x+\int x^2 d x -\int_0^3 x^3 d x+\int x^4 d x \ldots+C $</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/d3c6309d-6e32-4434-4b95-9816f59fb900/public"></p> </div>

### <Success>$\text{Expansion of} \ log (1-x)$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\therefore \log (1+x)= x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5} \cdots$</p> <p>To determine the value of $C$</p> <p>$\text { At } x =\log (1+x)=\log (1)=0$</p> <p>$ \therefore \quad C=0$</p> <p>By putting $x=-x$</p> <p>$ \log (1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4} \ldots $</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-16-infinite-series-l-4_5-hviteb7nlgk/img/310-1928.4.jpg"></p> </div>

### <Success>$\text{Expansion of} \ log (\cos x)$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\quad \log (\cos x) \quad \quad x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$</p> <p>$ f(x)=\log (\cos (x)) \quad \therefore f(0)=\log (1)=0$</p> <p>$ f^{\prime}(x)=\frac{1}{\cos x}(-\sin x)=-\tan x \quad\therefore f^{\prime}(0)=0$</p> <p>$f^{\prime\prime}(x)=-\left(1+\tan ^2 x\right)\quad \therefore f^{(2)}(0)=-1$</p> <p>$f^{\prime\prime\prime}(x)=-2 \tan x\left(1+\tan ^2\right)$</p> <p>$\quad\quad\quad=-2 \tan x-2 \tan ^3 x $</p> <p>$\therefore f^{\prime\prime\prime}(0)=0$</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-16-infinite-series-l-4_5-hviteb7nlgk/img/336-2082.0.jpg"></p> </div>

### <Success>$\text{Expansion of} \ log (\cos x)$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$ f^{(\prime\prime\prime\prime)}(x)=$</p> <p>$\quad-2\left(1+\tan ^2 x\right)-6 \tan ^2 x\left(1+\tan ^2 x\right)$</p> <p>$=-2-2 \tan ^2 x-6 \tan ^2 x-6 \tan ^4 x $</p> <p>$=-2-8 \tan ^2 x-6 \tan ^4 x$</p> <p>$ \therefore f^{(\prime\prime\prime\prime)} (0)=-2$</p> <p>$f^{(\prime\prime\prime\prime\prime)} (x)=-16(\tan x)\left(1+\tan ^2 x\right) -24 \tan ^3 x(1+\tan ^2 x) $</p> <p>$=-16 \tan x -16 \tan ^3 x -24 \tan ^3 x-24 \tan ^5 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-16-infinite-series-l-4_5-hviteb7nlgk/img/361-2241.6.jpg"></p> </div>

### <Success>$\text{Expansion of} \ log (\cos x)$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$ \therefore f^{(5)}(x)=-16 \tan x-40 \tan ^3 x-24 \tan ^5 x$</p> <p>$\therefore f^{(5)}(0)=0 $</p> <p>$\therefore f^{(6)}(x)=-16\left(1+\tan ^2\right)+\ldots $</p> <p>$\quad f^{(6)}(0)=-16$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/3d260c39-f4a2-4314-b0af-34be462d2900/public"></p> </div>

### <Success>$\text{Expansion of} \ log (\cos x)$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\therefore$ we find that for $\log (\cos (x))$</p> <p>$ f(0)=0$</p> <p>$ f^{\prime}(0)=0$</p> <p>$ f^{(2)}(0)=-1$</p> <p>$ f^{(3)}(0)=0$</p> <p>$ f^{(4)}(0)=-2$</p> <p>$ f^{(5)}(0)=0 $</p> <p>$ f^{(6)}(0)=-16$</p> </div> <div style='float: right; width:0%;'> </div>

### <Success>$\text{Expansion of} \ log (\cos x)$</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\therefore$ The series expansion is:</p> <p>$ -\frac{x^2}{2}-2 \frac{x^4}{4 !}-16 \frac{x^6}{6 !} \cdots$</p> <p>$ = -\frac{x^2}{2}-\frac{x^4}{12}-\frac{x^6}{45} \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-16-infinite-series-l-4_5-hviteb7nlgk/img/398-2472.0.jpg"></p> </div>

### <Success>Notes</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>We can also do it as follows:</p> <p>$log (\cos x) = log (1+(\cos x-1))$</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-16-infinite-series-l-4_5-hviteb7nlgk/img/419-2538.0.jpg"></p> </div>

### <Success>Euler's constant</Success> <div style='float: left; text-align:left width:60%;font-size:30px; '> <p>Euler’s constant “e”.</p> <p>$ \lim_{n \rightarrow \infty} (1+ \frac{1}{n})^n$</p> </div> <div style='float: right; width:40%;'> <img src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/bf0daf6c-f7eb-4a95-c37c-cfffe5cff400/public"> <!----> </div>

### <Success>Euler's constant</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>where does $\left(1+\frac{1}{x}\right)^x$ converge?</p> <p>$ n =2 \quad (1+0.5)^2=2.25$</p> <p>$ =10 \quad (1+0.1)^{10}=2.594$</p> <p>$ =100 \quad (1+0.01)^{100}=2.7048$</p> <p>$ =1000 \quad (1+0.001)^{1000}=2.7169$</p> <p>$ =10000 \quad (1+0.0001)^{10000}=2.7181$</p> <p>$ =100000 \quad(1+0.00001)^{100000}=2.7183$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/6058503d-d946-4d1d-94b0-450c3379f000/public"></p> </div>

### <Success>Euler's constant</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$ \text { As } n \rightarrow \infty$</p> <p>$ \left(1+\frac{1}{n}\right)^n \rightarrow 2 \cdot 718281828459045 \ldots$</p> <p>Euler’s Number.</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-16-infinite-series-l-4_5-hviteb7nlgk/img/509-3060.0.jpg"></p> </div>

### <Success>Question 2</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$ \left(1+\frac{1}{n}\right)^n $</p> <p>What is the $k^{\text {th }}$ term?</p> <p><strong>Solution:</strong></p> <p>For a given $k$ if we consider $n>k$</p> <p>$ \frac{n(n-1) \cdots(n-\overline{k-1})}{k!}(\frac{1}{n})^k $</p> <p>${^nc_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-16-infinite-series-l-4_5-hviteb7nlgk/img/522-3164.4.jpg"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align: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 !} $</p> <p>$\therefore$ As $x \rightarrow \infty$ the $k^{\text {th }}$ term in $ \frac{1}{k!}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/608138f1-14a4-412e-7413-6e6a5e6b3900/public"></p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left width:100%;font-size:30px;'> <p>$\therefore$ The series $\left(1+\frac{1}{n}\right)^n$</p> <p>$ \rightarrow 1+1+\frac{1}{2 !}+\frac{1}{3 !}+\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-16-infinite-series-l-4_5-hviteb7nlgk/img/539-3234.0.jpg"></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%;'> <p><img alt="image" src="http://115.241.75.180/iitpal-images/m11/mathematics-class-11-unit-10-chapter-16-infinite-series-l-4_5-hviteb7nlgk/img/541-3246.0.jpg"></p>