KC Sinha Solution Class 11 Chapter 16 कुछ महत्वपूर्ण अनंत श्रेणी (Some Important Infinite series) Exercise 16.5

 Exercise 16.5

निम्नलिखित कोे सिद्ध कीजिए 

(Prove the following)

Question 1

$\log _{e} 2-\frac{1}{2}=\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots$ to $\infty$

Sol :

$\log _{e} 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\cdots \infty$

$\log_e2=\frac{2-1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\dots \infty$

$\log _e{2}-\frac{1}{2}=\frac{1}{3}-\frac{1}{6}-\frac{1}{4}+\frac{1}{5}+\frac{1}{15}+\ldots \infty$

$=\frac{2-1}{6}-\frac{15+12+4}{60}+\dots \infty$

$=\frac{1}{6}+\frac{1}{60}+\ldots . \infty$

$\log _{e} 2-\frac{1}{2}=\frac{1}{1.2 .3}+\frac{1}{3.4 .5}+\dots \infty$

Question 2

$\frac{\log (1+x)}{x}=1-\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+\ldots$ to $\infty$

Sol :

L.H.S

$\frac{\log (1+2)}{x}=\frac{x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\dots ~to~\infty}{x}$

$=\frac{x\left(1-\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+\cdots \infty\right)}{x}$

$=1-\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+\cdots\infty$

Question 3

$\log \frac{1}{1+x}=-x+\frac{x^{2}}{2}-\frac{x^{3}}{3}+\frac{x^{4}}{4}-\ldots$ to $\infty$

Sol :

L.H.S

$\log \frac{1}{1+x}$=log 1 -log(1+x)

$=0-\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots \infty\right)$

$=-x+\frac{x^{2}}{2}-\frac{x^{3}}{3}+\frac{x^{4}}{4}+\cdots \infty$

Question 4

$1-\log 2=\frac{1}{2 \cdot 3}+\frac{1}{4 \cdot 5}+\frac{1}{6 \cdot 7}+\ldots$ to $\infty$

Sol :

$\log 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7} \cdots \infty$

$\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7} \ldots \infty=1-\log 2$

$\frac{3-2}{2.3}+\frac{5-4}{4.5}+\frac{7-6}{6.7}+\cdots \infty=1-\log 2$

$\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\cdots \infty=1-\log 2$

Question 7

log(n+1)-log(n-1)=2$\left[\frac{1}{n}+\frac{1}{3n^2}+\frac{1}{5n^5}+\dots\right]$

Sol :

L.H.S

=log(n+1)-log(n-1)

$=\log \frac{n+1}{n-1}$

$=\log \frac{\frac{n+1}{n}}{\frac{n-1}{n}}$

$=\log \frac{\left(\frac{n}{n}+\frac{1}{n}\right)}{\left(\frac{n}{n}-\frac{1}{n}\right)}$

$=\log \left(\frac{1+\frac{1}{n}}{1-\frac{1}{n}}\right)$

$\left[\log \left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}\dots \infty \right)\right}$

$=2\left[\frac{1}{n}+\frac{1}{3 n^{3}}+\frac{1}{5 n^{5}}+\cdots \infty\right]$

Question 11

2logx-log(x+1)-log(x-1)=$\frac{1}{x^{2}}+\frac{1}{2 x^{4}}+\frac{1}{3 x^{6}}+\ldots$

Sol :

L.H.S

=2logx-log(x+1)-log(x-1)

=log x²-[log(x+1)+log(x-1)]

=log x²-log(x+1)(x-1)

=log x²-log(x²-1²)

$=\log \frac{x^{2}}{x^{2}-1}$

$=\log \left(\frac{\frac{x^{2}}{x^{2}}}{\frac{x^{2}}{x^{2}}-\frac{1}{x^{2}}}\right)$

$=\log \left(\frac{1}{1-\frac{1}{x^{2}}}\right)$

$=\log \left(1-\frac{1}{x^{2}}\right)^{-1}$ $\left[-\log (1-x)=x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\dots \infty\right]$

$=-\log \left(1-\frac{1}{x^{2}}\right)=\frac{1}{x^{2}}+\left(\frac{1}{x^{2}}\right)^{2} \times \frac{1}{2}+\left(\frac{1}{x^{2}}\right)^{3} \times \frac{1}{3}+\dots\infty$

$=\frac{1}{x^{2}}+\frac{1}{2 x^{4}}+\frac{1}{3 x^{6}}+\ldots$