Question

Find whether the following series is convergent or divergent: $$x^2+{\displaystyle \frac{2^2}{3.4}}{x}^4+{\displaystyle \frac{2^2{4}^2}{3.4.5.6}}{x}^6+{\displaystyle \frac{2^2{4}^2{6}^2}{3.4.5.6.7.8}}{x}^8+...$$

Answer 1

Hint: A series is the sum of the terms of an infinite sequence of numbers. A series can be a convergent series or divergent series.
A series is a convergent series if the sequence of its partial sums tends to a limit; it means the partial sum becomes closer and closer to a given number when the number of their terms increases.
A series is a divergent series if the sequence of its partial sums does not tend to a finite limit; this series does not converge.
Use the Ratio test, which says if the ratio $$r<1$$series is convergent if $$r>1$$the series diverges, and if $$r=1$$the test fails, then use Raabe’s Test $$r=1$$ .

Complete step by step solution:
For the given series
 $$x^2+{\displaystyle \frac{2^2}{3.4}}{x}^4+{\displaystyle \frac{2^2{4}^2}{3.4.5.6}}{x}^6+{\displaystyle \frac{2^2{4}^2{6}^2}{3.4.5.6.7.8}}{x}^8+...$$
The sum of the series be
$$S_n=x^2+{\displaystyle \frac{2^2}{3.4}}{x}^4+{\displaystyle \frac{2^2{4}^2}{3.4.5.6}}{x}^6+{\displaystyle \frac{2^2{4}^2{6}^2}{3.4.5.6.7.8}}{x}^8+...$$
The $$T_n$$term of the series will be
$$T_n={\displaystyle \frac{2^2{4}^2{6}^2....{(2n)}^2{x}^{2n+2}}{3.4.5.6.7.8......\left(2n+1\right)\left(2n+2\right)}}$$
Hence the $$T_{n+1}$$term of the series will be
$$T_{n+1}={\displaystyle \frac{2^2{4}^2{6}^2....{(2n)}^2{(2n+2)}^2{x}^{2n+4}}{3.4.5.6.7.8......\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)\left(2n+4\right)}}$$
Now use the ratio test for the given infinite series, so
$$\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{T_{n+1}}{T_n}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\left(2n+2\right)}^2{x}^2}{\left(2n+3\right)\left(2n+4\right)}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\left(2+{\displaystyle \frac{2}{n}}\right)}^2{x}^2}{\left(2+{\displaystyle \frac{3}{n}}\right)\left(2+{\displaystyle \frac{4}{n}}\right)}}$$
Now put the value of$$n\to \mathrm{\infty }$$, we get
 $$\begin{gathered} \underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{T_{n+1}}{T_n}}=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\left(2n+2\right)}^2{x}^2}{\left(2n+3\right)\left(2n+4\right)}} \\ =\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\left(2+{\displaystyle \frac{2}{n}}\right)}^2{x}^2}{\left(2+{\displaystyle \frac{3}{n}}\right)\left(2+{\displaystyle \frac{4}{n}}\right)}} \\ =\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\left(2+{\displaystyle \frac{2}{\mathrm{\infty }}}\right)}^2{x}^2}{\left(2+{\displaystyle \frac{3}{\mathrm{\infty }}}\right)\left(2+{\displaystyle \frac{4}{\mathrm{\infty }}}\right)}} \\ ={\displaystyle \frac{{\left(2+0\right)}^2{x}^2}{\left(2+0\right)\left(2+0\right)}} \\ =x^2 \end{gathered}$$
Now for the value of$$x>1$$, $$S_n$$diverges
For$$-1<x<1$$, $$S_n$$converges
And if$$x=1$$the ratio test fails
As already explained, if the ratio test fails, we use Raabe’s Test
Raabe’s test is given by the formula $$\underset{n\to \mathrm{\infty }}{lim}\left[{\displaystyle \frac{T_n}{T_{n+1}}}-1\right]n$$; hence by applying in this series where$$x=1$$we get
$$\begin{gathered} \underset{n\to \mathrm{\infty }}{lim}\left[{\displaystyle \frac{T_n}{T_{n+1}}}-1\right]n=\underset{n\to \mathrm{\infty }}{lim}\left[{\displaystyle \frac{\left(2n+3\right)\left(2n+4\right)}{{\left(2n+2\right)}^2}}-1\right]n \\ =\underset{n\to \mathrm{\infty }}{lim}\left[{\displaystyle \frac{4n^2+8n+6n+12}{4n^2+8n+4}}-1\right]n \\ =\underset{n\to \mathrm{\infty }}{lim}\left[{\displaystyle \frac{4n^2+14n+12-4n^2-8n-4}{4n^2+8n+4}}\right]n \\ =\underset{n\to \mathrm{\infty }}{lim}n\left[{\displaystyle \frac{6n+8}{4n^2+8n+4}}\right] \\ =\underset{n\to \mathrm{\infty }}{lim}\left[{\displaystyle \frac{6n^2+8n}{4n^2+8n+4}}\right] \end{gathered}$$
Now taking common from both the numerator and the denominator
$$\begin{gathered} \underset{n\to \mathrm{\infty }}{lim}\left[{\displaystyle \frac{T_n}{T_{n+1}}}-1\right]n=\underset{n\to \mathrm{\infty }}{lim}\left[{\displaystyle \frac{6n^2+8n}{4n^2+8n+4}}\right] \\ =\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{n^2}{n^2}}\left[{\displaystyle \frac{6+{\displaystyle \frac{8}{n}}}{4+{\displaystyle \frac{8}{n}}+{\displaystyle \frac{4}{n^2}}}}\right] \end{gathered}$$
Now put the value of$$n\to \mathrm{\infty }$$, we get
$$\begin{gathered} \underset{n\to \mathrm{\infty }}{lim}\left[{\displaystyle \frac{T_n}{T_{n+1}}}-1\right]n=\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{n^2}{n^2}}\left[{\displaystyle \frac{6+{\displaystyle \frac{8}{n}}}{4+{\displaystyle \frac{8}{n}}+{\displaystyle \frac{4}{n^2}}}}\right] \\ =\underset{n\to \mathrm{\infty }}{lim}\left[{\displaystyle \frac{6+{\displaystyle \frac{8}{\mathrm{\infty }}}}{4+{\displaystyle \frac{8}{\mathrm{\infty }}}+{\displaystyle \frac{4}{{\mathrm{\infty }}^2}}}}\right] \\ =\left[{\displaystyle \frac{6+0}{4+0+0}}\right] \\ ={\displaystyle \frac{6}{4}} \\ ={\displaystyle \frac{3}{2}} \end{gathered}$$
Since the value of $$n={\displaystyle \frac{3}{2}}>1$$
Hence the series is convergent at $$x=1$$

Note: Students must remember that if a number is divided by an infinite value, then the result will always be equal to 0. Moreover, if the ratio test fails then, it does not ensure the nature of the series.