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

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.