Question

How do you test the series \[\sum{\dfrac{{{n}^{2}}}{{{2}^{n}}}}\] from n is \[[0,\infty )\] for convergence?

Answer 1

Hint: We say that a series is convergent if the summation of the series from the initial starting value to the final ending value becomes equal to some constant value or may approach some constant value under some constraints. We have to keep in mind of the different tests we can perform to test for the convergence of a series. The above problem can be done easily by the Ratio test or the d’Alembert’s ratio test. The test states that,
If \[L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|\] , and if,
1. \[L<1\] , then the series is convergent.
2. \[L>1\], then the series is divergent.
3. \[L=1\] , then the series may or may not converge and we need to perform other tests.

Complete step by step answer:
Now, starting off the problem, we can write,
\[{{a}_{n}}=\dfrac{{{n}^{2}}}{{{2}^{n}}}\] and \[{{a}_{n+1}}=\dfrac{{{\left( n+1 \right)}^{2}}}{{{2}^{n+1}}}\]
We now apply the d’Alembert’s ratio test to the given problem and check for the value of \[L\] .
We write \[L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|\] . Now putting the values in the respective positions we get,
\[L=\displaystyle \lim_{n \to \infty }\left| \dfrac{\dfrac{{{\left( n+1 \right)}^{2}}}{{{2}^{n+1}}}}{\dfrac{{{n}^{2}}}{{{2}^{n}}}} \right|\]
Now, multiplying \[{{2}^{n}}\] in both the numerator and denominator we get,
\[L=\displaystyle \lim_{n \to \infty }\left| \dfrac{\dfrac{{{\left( n+1 \right)}^{2}}}{{{2}^{n+1}}}\times {{2}^{n}}}{\dfrac{{{n}^{2}}}{{{2}^{n}}}\times {{2}^{n}}} \right|\]
On cancelling the common terms we get,
\[L=\displaystyle \lim_{n \to \infty }\left| \dfrac{\dfrac{{{\left( n+1 \right)}^{2}}}{2}}{{{n}^{2}}} \right|\]
Now, expanding the numerator we get,
\[L=\displaystyle \lim_{n \to \infty }\left| \dfrac{\dfrac{{{n}^{2}}+2n+1}{2}}{{{n}^{2}}} \right|\] ,
We now divide both the numerator and denominator by \[{{n}^{2}}\] and we get,
\[L=\displaystyle \lim_{n \to \infty }\left| \dfrac{\dfrac{\dfrac{{{n}^{2}}}{{{n}^{2}}}+\dfrac{2n}{{{n}^{2}}}+\dfrac{1}{{{n}^{2}}}}{2}}{\dfrac{{{n}^{2}}}{{{n}^{2}}}} \right|\] ,
Now cancelling the common terms we get,
\[L=\displaystyle \lim_{n \to \infty }\left| \dfrac{1+\dfrac{2}{n}+\dfrac{1}{{{n}^{2}}}}{2} \right|\] ,
Now, from the knowledge of limit theory, we can write,
\[\displaystyle \lim_{n \to \infty }\dfrac{2}{n}=0\] and \[\displaystyle \lim_{n \to \infty }\dfrac{1}{{{n}^{2}}}=0\] .
Now, putting these two values in the original equation we get,
\[L=\displaystyle \lim_{n \to \infty }\left| \dfrac{1+0+0}{2} \right|\] ,
\[\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{1}{2} \right|\]
\[\Rightarrow L=\dfrac{1}{2}\] ,
Now, from the rules of the test we know that if \[L<1\] the series converges. From the solution we can clearly see that \[\dfrac{1}{2}<1\] , and thus our given series converges.

Note: We need to keep in mind all the different tests for convergence of series and tactically think of the one that suits our given problem and apply them accordingly. However the above problem can also be done using the other tests as well, but this solution is a simpler one to do and understand. Some of the different important tests include Ratio test or the d’Alembert’s ratio test, Cauchy’s root test, the integral test, the Abel’s test.