(1 point) The following series are geometric series or a sum...

(1 point) The following series are geometric series or a sum of two geometric series. Determine whether each series converges or not. For the series which converge, enter the sum of the series. For the series which diverges enter "DIV" (without quotes). (a) sum_{n=1}^{infinity} 5^n / 4^n = DIV (b) sum_{n=2}^{infinity} 1 / 2^n = 1/2 (c) sum_{n=0}^{infinity} 2^n / 9^{2n+1} = 0 (d) sum_{n=5}^{infinity} 4^n / 5^n = 0 (e) sum_{n=1}^{infinity} 6^n / 6^{n+4} = DIV (f) sum_{n=1}^{infinity} (4^n + 2^n) / 5^n = 14/3

Transcript

0:00 Hi.

00:02 Now given the series, summation n is equal to 1 to infinity, 5 to the power n by 4 to the power n.

00:11 So this can be written as submission n is equal to 1 to infinity 5 by 4 whole to the power n.

00:18 So this can be written as 5 by 4 plus 5 by 4 whole square plus 5 by 4 whole cube and so on.

00:29 So, this can further be written as 5 by 4, 1 plus 5 by 4 plus 5 by 4 whole square plus so on.

00:40 Now comparing this with geometric series a 1 plus r plus r square plus so on.

00:50 So here we have a is equal to 5 by 4 and r is equal to 5 by 4.

00:56 Now my mod r is mod 5 by 4 which is greater than 1.

01:02 So my series is divergent.

01:06 Next we have summation n is equal to 2 to infinity 1 by 2 to the power n.

01:15 So this can be written as 1 by 2 square plus 1 by 2 cube plus 1 by 2 to the power 4 and so on.

01:26 So taking 1 by 2 square common, we have 1 plus 1 by 2 plus 1 by 2 square and so on.

01:37 So from here we see that my a is 1 by 2 square which is 1 by 4 and my r is 1 by 2 which is less than 1.

01:49 So my series is convergent and for this sum is a by 1.

01:56 Minus r which is equal to 1 by 4 by 1 minus half which comes out to be half next we have submission n is equal to 0 to infinity 2 to the power n by 9 to the power 2 n plus 1 so by this series becomes 1 by 9 plus 2 by 9 cube plus 4 by 9 to the power 5 and so on so now this can further be written as 1 by 9 common 1 plus 2 by 9 square plus 4 by 9 to the power 4 and so on.

02:43 So from here my a is 1 by 9 and my r is 2 by 9 square.

02:50 So mod r is less than 1.

02:52 So my series is convergent and sum is a by 1.

02:58 Minus r.

03:00 A is 1 by 9 by 1 minus 2 by 81...

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