Question

Definite Integral as the Limit of a Sum

Answer 1

Hint: We recall the definition of power series otherwise known as Taylor’s series and convergence. We use the standard Taylor’s series for the function $ \ln \left( 1+x \right) $ centred around $ x=a $ which is given by $ \ln \left( 1+x \right)=\ln \left( 1+a \right)+\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( x-a \right)}^{n}}{{\left( -1 \right)}^{n+1}}}{n{{\left( 1+a \right)}^{n}}}} $ . We put $ {{x}^{2}} $ instead of $ x $ to get the power series of $ f\left( x \right)=\ln \left( {{x}^{2}}+1 \right) $ and radius of convergence $ r $ such that $ \left| x \right| < r $ obtained from the condition of convergence $ \displaystyle \lim_{n \to \infty }\left| \dfrac{{{t}_{n+1}}}{{{t}_{n}}} \right| < 1 $ .\[\]

Complete step by step answer:
We know that power series are given with infinite terms
\[\sum\limits_{n=0}^{\infty }{{{a}_{n}}{{\left( x-c \right)}^{n}}}={{a}_{0}}+{{a}_{1}}\left( x-c \right)+{{a}_{2}}{{\left( x-c \right)}^{2}}+...\]
Here $ n $ is the power and $ c $ is called centre. We say is series is convergent around the centre $ c $ if $ \left| r \right|
 < 1 $ where
\[r=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{t}_{n+1}}}{{{t}_{n}}} \right|\]
 We also know that we can approximate any infinitely differentiable function $ f\left( x \right) $ as power series with centre $ x=a $ using Taylor’s approximation formula as written below
\[f\left( x \right)=\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{n}}\left( a \right)\left( x-a \right)}{n!}}\]
We know the Taylor’s series approximation of natural logarithmic function $ \ln \left( 1+x \right) $ around the centre $ x=a $ as
\[\ln \left( 1+x \right)=\ln \left( 1+a \right)+\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( x-a \right)}^{n}}{{\left( -1 \right)}^{n+1}}}{n{{\left( 1+a \right)}^{n}}}}\]
We are asked in the question to find the Taylor’s series approximation of $ \ln \left( {{x}^{2}}+1 \right)=\ln \left( 1+{{x}^{2}} \right) $ which is similar to the power series of $ \ln \left( 1+x \right) $ . We put $ {{x}^{2}} $ instead of $ x $ in the above power series of $ \ln \left( 1+x \right) $ and the required power series around the centre $ x=a $ as
\[\ln \left( 1+{{x}^{2}} \right)=\ln \left( 1+a \right)+\sum\limits_{n=1}^{\infty }{\dfrac{{{\left( {{x}^{2}}-a \right)}^{n}}{{\left( -1 \right)}^{n+1}}}{n{{\left( 1+a \right)}^{n}}}}\]
We shall now apply the ratio test for testing convergence of the above series. We take the ratio of $ {{\left( n+1 \right)}^{th}} $ and $ {{n}^{th}} $ term and take limit $ n \to \infty $ to have
\[\begin{align}
  & \displaystyle \lim_{n \to \infty }\left| \dfrac{{{t}_{n+1}}}{{{t}_{n}}} \right|=\displaystyle \lim_{n \to \infty }\left| \dfrac{\dfrac{{{\left( {{x}^{2}}-a \right)}^{n+1}}{{\left( -1 \right)}^{n+1+1}}}{\left( n+1 \right){{\left( 1+a \right)}^{n+1}}}}{\dfrac{{{\left( {{x}^{2}}-a \right)}^{n}}{{\left( -1 \right)}^{n+1}}}{n{{\left( 1+a \right)}^{n}}}} \right| \\
 & \Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{t}_{n+1}}}{{{t}_{n}}} \right|=\displaystyle \lim_{n \to \infty }\left| \dfrac{\left( {{x}^{2}}-a \right)}{\left( 1+a \right)\left( n+1 \right)}\times n \right| \\
 & \Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{t}_{n+1}}}{{{t}_{n}}} \right|=\left| \dfrac{{{x}^{2}}-1}{1+a} \right|\displaystyle \lim_{n \to \infty }\left| \dfrac{n}{n+1} \right| \\
 & \Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{{{t}_{n+1}}}{{{t}_{n}}} \right|=\left| \dfrac{{{x}^{2}}-1}{1+a} \right|\times 1=\left| \dfrac{{{x}^{2}}-1}{1+a} \right| \\
\end{align}\]
So the absolute ratio has to be less than1 for the series to converge. We have;
\[\begin{align}
  & \left| \dfrac{{{x}^{2}}-a}{a+1} \right| < 1 \\
 & \Rightarrow \left| {{x}^{2}}-a \right| < a+1 \\
 & \Rightarrow \left| {{x}^{2}} \right| < 2a+1 \\
 & \Rightarrow \left| x \right| < \sqrt{1+2a} \\
\end{align}\]
So the radius of the convergence is $ \sqrt{1+2a} $ and convergence depends upon the centre $ x=a $ . \[\]

Note:
We note that the if we take centre $ x=0 $ we get the Maclaurine’s series as $ \sum\limits_{n=0}^{\infty }{{{a}_{n}}{{x}^{n}}}={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+... $ Here in this problem if we take centre $ x=0 $ then we get the alternating series $ \ln \left( 1+{{x}^{2}} \right)=\sum\limits_{n=1}^{\infty }{\dfrac{{{x}^{2n}}{{\left( -1 \right)}^{n+1}}}{n}}={{x}^{2}}-\dfrac{{{x}^{2}}}{4}+\dfrac{{{x}^{6}}}{3}-... $ . The Taylor’s series is used to solve the definite integral problem where the functions are non-integrable.