Respuesta :
The given series [tex]\sum_{n=1}^{\infty}\left [ (0.3)^{n-1}-(0.2)^{n} \right ][/tex] is convergence and the sum of the series is 33/28.
In the given question we have to check whether the series converge or not. If the series does converge, then we have to find the sum.
The given series is [tex]\sum_{n=1}^{\infty}\left [ (0.3)^{n-1}-(0.2)^{n} \right ][/tex].
Firstly we check the convergence of the given series.
When a series approaches a limit, it is considered to be convergent if both convergent and divergent behaviour occur. The deletion of a finite number of terms from the beginning of a series has no impact on convergence or divergence.
We can write the series as:
[tex]\sum_{n=1}^{\infty}\left [ (0.3)^{n-1}-(0.2)^{n} \right ]=\sum_{n=1}^{\infty} (0.3)^{n-1}-\sum_{n=1}^{\infty}(0.2)^{n} \right[/tex]
As |0.3|<1 and |0.2|<1
Then [tex]\sum_{n=1}^{\infty}(0.3)^{n-1}\text{ and }\sum_{n=1}^{\infty}(0.2)^n[/tex] is convergent.
So [tex]\sum_{n=1}^{\infty} (0.3)^{n-1}-\sum_{n=1}^{\infty}(0.2)^{n} \right[/tex] is also convergent.
[tex]\sum_{n=1}^{\infty}\left [ (0.3)^{n-1}-(0.2)^{n} \right ]=\sum_{n=1}^{\infty} (0.3)^{n-1}-(0.2)\sum_{n=1}^{\infty}(0.2)^{n-1} \right[/tex]
Series [tex]\sum_{n=1}^{\infty}(x)^{n-1}[/tex] converges to 1/x-1 for |x|<1.
[tex]\sum_{n=1}^{\infty}\left [ (0.3)^{n-1}-(0.2)^{n} \right ]=\frac{1}{1-0.3}-(0.2)\frac{1}{1-0.2}[/tex]
[tex]\sum_{n=1}^{\infty}\left [ (0.3)^{n-1}-(0.2)^{n} \right ]=\frac{1}{0.7}-\frac{0.2}{0.8}[/tex]
[tex]\sum_{n=1}^{\infty}\left [ (0.3)^{n-1}-(0.2)^{n} \right ][/tex] = 10/7 - 1/4
[tex]\sum_{n=1}^{\infty}\left [ (0.3)^{n-1}-(0.2)^{n} \right ][/tex] = 33/28
Hence, the sum of the given series is 33/28.
To learn more about the convergence of series link is here
brainly.com/question/15415793
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The right question is:
Determine whether the series is convergent or divergent.
[tex]\sum_{n=1}^{\infty}\left [ (0.3)^{n-1}-(0.2)^{n} \right ][/tex]
If is is convergent, find the sum. (If the quantity diverges, enter diverges.)
