Question

How do you determine if ${{a}_{n}}={{\left( 1+\dfrac{1}{{{n}^{2}}} \right)}^{n}}$ converge and find the limits when they exist?

Answer 1

We start solving the problem by finding the values of n at which the given series is not valid. We then take the limit of the function at these values by first applying logarithm on both sides of the series. We then make the necessary calculations and make use of the L-Hospital rule to proceed through the problem. We then make the necessary calculations to get the required value of limit which tells us whether the series is convergent or divergent.

Complete step by step answer:
According to the problem, we are asked to determine whether the given series ${{a}_{n}}={{\left( 1+\dfrac{1}{{{n}^{2}}} \right)}^{n}}$ converge and we need to find the limits (if they exist).
We know that in order to check whether the given series is convergent or divergent, we need to check them at the values for which the series is not valid which are $n=0$ and $n=\infty$.
Now, let us find the limit $\displaystyle \lim_{n\to 0}{{a}_{n}}=\displaystyle \lim_{n\to 0}{{\left( 1+\dfrac{1}{{{n}^{2}}} \right)}^{n}}$.
$\Rightarrow \displaystyle \lim_{n\to 0}{{a}_{n}}={{\left( 1+\dfrac{1}{{{0}^{2}}} \right)}^{0}}={{\infty }^{0}}$, which is an indeterminate form.
Now, let us apply logarithm on both sides of ${{a}_{n}}={{\left( 1+\dfrac{1}{{{n}^{2}}} \right)}^{n}}$.
So, we have $\log \left( {{a}_{n}} \right)=\log {{\left( 1+\dfrac{1}{{{n}^{2}}} \right)}^{n}}$ ---(1).
We know that $\log {{x}^{a}}=a\log x$. Let us use this result in equation (1).
$\Rightarrow \log \left( {{a}_{n}} \right)=n\log \left( 1+\dfrac{1}{{{n}^{2}}} \right)$.
$\Rightarrow \log \left( {{a}_{n}} \right)=\dfrac{\log \left( 1+\dfrac{1}{{{n}^{2}}} \right)}{\dfrac{1}{n}}$ ---(2).
Now, let us apply limits on both sides of equation (2).
$\Rightarrow \displaystyle \lim_{n\to 0}\log \left( {{a}_{n}} \right)=\displaystyle \lim_{n\to 0}\dfrac{\log \left( 1+\dfrac{1}{{{n}^{2}}} \right)}{\dfrac{1}{n}}=\dfrac{\log \left( 1+\dfrac{1}{{{0}^{2}}} \right)}{\dfrac{1}{0}}=\dfrac{\infty }{\infty }$, which is an indeterminate form.
So, we can apply the L-Hospital rule. We know that L-hospital is defined as if limit of any function results out to be $\dfrac{\infty }{\infty }$, $\dfrac{0}{0}$, then $\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{{{f}^{'}}\left( x \right)}{{{g}^{'}}\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{{{f}^{''}}\left( x \right)}{{{g}^{''}}\left( x \right)}=......$.
$\Rightarrow \displaystyle \lim_{x \to 0}\log \left( {{a}_{n}} \right)=\displaystyle \lim_{x \to 0}\dfrac{\dfrac{d}{dn}\left( \log \left( 1+\dfrac{1}{{{n}^{2}}} \right) \right)}{\dfrac{d}{dn}\left( \dfrac{1}{n} \right)}$.
$\Rightarrow \displaystyle \lim_{n\to 0}\log \left( {{a}_{n}} \right)=\displaystyle \lim_{n\to 0}\dfrac{\dfrac{\dfrac{-2}{{{n}^{3}}}}{\left( \dfrac{{{n}^{2}}+1}{{{n}^{2}}} \right)}}{\dfrac{-1}{{{n}^{2}}}}$.
$\Rightarrow \displaystyle \lim_{n\to 0}\log \left( {{a}_{n}} \right)=\displaystyle \lim_{n\to 0}\dfrac{\dfrac{-2}{n\left( {{n}^{2}}+1 \right)}}{\dfrac{-1}{{{n}^{2}}}}$.
$\Rightarrow \displaystyle \lim_{n\to 0}\log \left( {{a}_{n}} \right)=\displaystyle \lim_{n\to 0}\dfrac{2{{n}^{2}}}{{{n}^{3}}+n}=\dfrac{0}{0}$. So, let us apply the L-Hospital rule again.
$\Rightarrow \displaystyle \lim_{n\to 0}\log \left( {{a}_{n}} \right)=\displaystyle \lim_{n\to 0}\dfrac{\dfrac{d}{dn}\left( 2{{n}^{2}} \right)}{\dfrac{d}{dn}\left( {{n}^{3}}+n \right)}$.
$\Rightarrow \displaystyle \lim_{n\to 0}\log \left( {{a}_{n}} \right)=\displaystyle \lim_{n\to 0}\dfrac{4n}{3{{n}^{2}}+1}=\dfrac{0}{1}=0$.
$\Rightarrow \displaystyle \lim_{n\to 0}{{a}_{n}}={{e}^{0}}=1$.
So, the given function is convergent at $n=0$.
Now, let us find the limit $\displaystyle \lim_{n\to \infty }{{a}_{n}}=\displaystyle \lim_{n\to \infty }{{\left( 1+\dfrac{1}{{{n}^{2}}} \right)}^{n}}$.
$\Rightarrow \displaystyle \lim_{n\to \infty }{{a}_{n}}={{\left( 1+\dfrac{1}{{{\infty }^{2}}} \right)}^{\infty }}={{1}^{\infty }}$, which is an indeterminate form.
Now, let us apply logarithm on both sides of ${{a}_{n}}={{\left( 1+\dfrac{1}{{{n}^{2}}} \right)}^{n}}$.
So, we have $\log \left( {{a}_{n}} \right)=\log {{\left( 1+\dfrac{1}{{{n}^{2}}} \right)}^{n}}$ ---(3).
We know that $\log {{x}^{a}}=a\log x$. Let us use this result in equation (3).
$\Rightarrow \log \left( {{a}_{n}} \right)=n\log \left( 1+\dfrac{1}{{{n}^{2}}} \right)$.
$\Rightarrow \log \left( {{a}_{n}} \right)=\dfrac{\log \left( 1+\dfrac{1}{{{n}^{2}}} \right)}{\dfrac{1}{n}}$ ---(2).
Now, let us apply limits on both sides of equation (2).
$\Rightarrow \displaystyle \lim_{n\to \infty }\log \left( {{a}_{n}} \right)=\displaystyle \lim_{n\to \infty }\dfrac{\log \left( 1+\dfrac{1}{{{n}^{2}}} \right)}{\dfrac{1}{n}}=\dfrac{\log \left( 1+\dfrac{1}{{{\infty }^{2}}} \right)}{\dfrac{1}{\infty }}=\dfrac{0}{0}$, which is an indeterminate form.
So, we can apply the L-Hospital rule.
$\Rightarrow \displaystyle \lim_{x \to \infty }\log \left( {{a}_{n}} \right)=\displaystyle \lim_{x \to \infty }\dfrac{\dfrac{d}{dn}\left( \log \left( 1+\dfrac{1}{{{n}^{2}}} \right) \right)}{\dfrac{d}{dn}\left( \dfrac{1}{n} \right)}$.
$\Rightarrow \displaystyle \lim_{n\to \infty }\log \left( {{a}_{n}} \right)=\displaystyle \lim_{n\to \infty }\dfrac{\dfrac{\dfrac{-2}{{{n}^{3}}}}{\left( \dfrac{{{n}^{2}}+1}{{{n}^{2}}} \right)}}{\dfrac{-1}{{{n}^{2}}}}$.
$\Rightarrow \displaystyle \lim_{n\to \infty }\log \left( {{a}_{n}} \right)=\displaystyle \lim_{n\to \infty }\dfrac{\dfrac{-2}{n\left( {{n}^{2}}+1 \right)}}{\dfrac{-1}{{{n}^{2}}}}$.
$\Rightarrow \displaystyle \lim_{n\to \infty }\log \left( {{a}_{n}} \right)=\displaystyle \lim_{n\to \infty }\dfrac{2{{n}^{2}}}{{{n}^{3}}+n}$.
$\Rightarrow \displaystyle \lim_{n\to \infty }\log \left( {{a}_{n}} \right)=\displaystyle \lim_{n\to \infty }\dfrac{2}{n+\dfrac{1}{n}}$.
$\Rightarrow \displaystyle \lim_{n\to \infty }\log \left( {{a}_{n}} \right)=\displaystyle \lim_{n\to \infty }\dfrac{1}{n}\times \dfrac{2}{\left( 1+\dfrac{1}{{{n}^{2}}} \right)}$.
$\Rightarrow \displaystyle \lim_{n\to \infty }\log \left( {{a}_{n}} \right)=\dfrac{1}{\infty }\times \dfrac{2}{\left( 1+\dfrac{1}{{{\infty }^{2}}} \right)}=0\times 2=0$.
$\Rightarrow \displaystyle \lim_{n\to \infty }{{a}_{n}}={{e}^{0}}=1$.
So, the given function is convergent at $n=\infty$.

Note:
We can see that the given problem contains a huge amount of calculation, so we need to perform each step carefully to avoid confusion and calculation mistakes. Whenever we get the limit in the indeterminate forms like ${{0}^{0}}$, ${{\infty }^{0}}$, ${{\infty }^{\infty }}$, ${{1}^{\infty }}$, ${{0}^{\infty }}$ then we need to apply logarithm on both sides to get the limit. We should not report ${{1}^{\infty }}$ as 1, which is a common mistake one by students. Similarly, we can expect problems to find the continuity of the series at $n=0$.