Question

How do you find the power series for \[f\left( x \right)=\int{\dfrac{\ln \left( 1+t \right)}{t}}dt~\] from \[\left[ 0,x \right]\] and determine its radius of convergence?

Answer 1

Hint: In order to find the solution to the given question, that is to find the power series for \[f\left( x \right)=\int{\dfrac{\ln \left( 1+t \right)}{t}}dt~\] from \[\left[ 0,x \right]\] and determine its radius of convergence, we will use the Maclaurin Series of \[\ln \left( 1+x \right)\] then replace \[x\] with \[t\] to determine the value of \[\dfrac{\ln \left( 1+t \right)}{t}\] and find the power series with the help of it. And to determine radius of convergence use the d'Alembert's ratio test.

Complete step by step solution:
According to the question, given function in the question is as follows:
\[f\left( x \right)=\int{\dfrac{\ln \left( 1+t \right)}{t}}dt~\]
We are going to start with the well-known Maclaurin Series for \[\ln \left( 1+x \right)\]:
\[\ln \left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+...\]
Now replace the variable \[x\] in the above series by \[t\] then we get:
\[\Rightarrow \ln \left( 1+t \right)=t-\dfrac{{{t}^{2}}}{2}+\dfrac{{{t}^{3}}}{3}-\dfrac{{{t}^{4}}}{4}+...\]
So, now we have the value of \[\dfrac{\ln \left( 1+t \right)}{t}\] as:
\[\Rightarrow \dfrac{\ln \left( 1+t \right)}{t}=\dfrac{1}{t}\left\{ t-\dfrac{{{t}^{2}}}{2}+\dfrac{{{t}^{3}}}{3}-\dfrac{{{t}^{4}}}{4}+... \right\}\]
\[\Rightarrow \dfrac{\ln \left( 1+t \right)}{t}=1-\dfrac{t}{2}+\dfrac{{{t}^{2}}}{3}-\dfrac{{{t}^{3}}}{4}+...\]
So, we can substitute the above value in the given function and get:
\[\Rightarrow f\left( x \right)=\int\limits_{0}^{x}{\dfrac{\ln \left( 1+t \right)}{t}}dt\]
\[\Rightarrow f\left( x \right)=\int\limits_{0}^{x}{\left\{ 1-\dfrac{t}{2}+\dfrac{{{t}^{2}}}{3}-\dfrac{{{t}^{3}}}{4}+... \right\}}dt\]
\[\Rightarrow f\left( x \right)=\left[ t-\dfrac{\dfrac{{{t}^{2}}}{2}}{2}+\dfrac{\dfrac{{{t}^{3}}}{3}}{3}-\dfrac{\dfrac{{{t}^{4}}}{4}}{4}+... \right]_{0}^{x}\]
\[\Rightarrow f\left( x \right)=\left[ t-\dfrac{{{t}^{2}}}{4}+\dfrac{{{t}^{3}}}{9}-\dfrac{{{t}^{4}}}{16}+... \right]_{0}^{x}\]
\[\Rightarrow f\left( x \right)=x-\dfrac{{{x}^{2}}}{4}+\dfrac{{{x}^{3}}}{9}-\dfrac{{{x}^{4}}}{16}+...\]
Therefore, we can write the above equation in the following power series
\[\Rightarrow f\left( x \right)=\sum\limits_{r=1}^{\infty }{{{\left( -1 \right)}^{r+1}}\dfrac{{{x}^{r}}}{{{r}^{2}}}}\]
In order to find the radius of convergence of the derived power series we will apply
d'Alembert's ratio test which states that
Suppose that;
\[S=\sum\limits_{r=1}^{\infty }{{{a}_{n}}}\] and \[L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|\]
Then
if \[L<1\] then the series converges absolutely;
if \[L>1\] then the series is divergent;
if \[L=1\]or the limit fails to exist the test is inconclusive.

Now the given series is as follows;
\[\Rightarrow S=\sum\limits_{r=1}^{\infty }{{{\left( -1 \right)}^{r+1}}\dfrac{{{x}^{r}}}{{{r}^{2}}}}\]
Applying the d'Alembert's ratio test in the above series we get:
\[\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{\dfrac{{{\left( -1 \right)}^{n+1+1}}{{x}^{n+1}}}{{{\left( n+1 \right)}^{2}}}}{\dfrac{{{\left( -1 \right)}^{n+1}}{{x}^{n}}}{{{n}^{2}}}} \right|\]
After simplifying it further we get:
\[\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{\left( -1 \right)x{{n}^{2}}}{{{\left( n+1 \right)}^{2}}} \right|\]
Now open the brackets and solve them, we get:
\[\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| -x \right|\times \left| \dfrac{{{n}^{2}}}{{{n}^{2}}+2n+1}\cdot \dfrac{\dfrac{1}{{{n}^{2}}}}{\dfrac{1}{{{n}^{2}}}} \right|\]
\[\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| -x \right|\times \left| \dfrac{1}{1+\dfrac{2}{n}+\dfrac{1}{{{n}^{2}}}} \right|\]
After applying the limit, we get:
\[\Rightarrow L=\left| x \right|\]
And we can conclude that the series converges if \[L<1\]
\[\Rightarrow \left| x \right|<1\].
Therefore, \[f\left( x \right)=x-\dfrac{{{x}^{2}}}{4}+\dfrac{{{x}^{3}}}{9}-\dfrac{{{x}^{4}}}{16}+...\] or we can say \[f\left( x \right)=\sum\limits_{r=1}^{\infty }{{{\left( -1 \right)}^{r+1}}\dfrac{{{x}^{r}}}{{{r}^{2}}}}\] and its radius of convergence is \[\left| x \right|<1\] or equivalently \[x\in \left( -1,1 \right)\] or \[-1< x< 1\].

Note: Students make mistakes while getting confused with the signs before the terms in the Maclaurin series of \[\ln \left( 1+x \right)\]. They might write it as \[\ln \left( 1+x \right)=x+\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{4}}}{4}+...\] instead of \[\ln \left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+...\] which leads to the completely wrong answer.