(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) ?_{n=1}^? 9^n / 8^n = DIV (b) ?_{n=2}^? 1 / 2^n = 1/2 (c) ?_{n=0}^? 2^n / 9^{2n+1} = 2/81 (d) ?_{n=5}^? 8^n / 9^n = 8^5/9^4 (e) ?_{n=1}^? 6^n / 6^{n+4} = DIV (f) ?_{n=1}^? (8^n + 2^n) / 9^n = 58/7

Transcript

00:01 Hi, given a geometric series, we have to tell whether the series converges or diverges, and if it converges, we have to find the sum.

00:09 Now, a geometric series can be represented by a, a, r, a, r square and so on.

00:17 So this can be written as a1 plus r plus r square and so on.

00:23 Now, a geometric series converges if mod r is less than 1, and in this case, sum is equal to a, by 1 minus r and the series diverges if mod r is greater than 1.

00:36 So now given over here the series summation n is equal to 1 to infinity, 9 by 8 to the power n.

00:44 So this is equal to 9 by 8 plus 9 by 8 whole square plus 9 by 8 whole cube plus so on.

00:54 So from here taking 9 by 8 common we have 1 plus 9 by 8 plus 9 by 8 plus 9 by 1.

01:00 8 whole square and so on so from here we have a is equal to 9 by 8 and r is equal to 9 by 8 so we see that mod r over here is greater than 1 so the series over here diverges next we are given the series as submission n is equal to 2 to infinity 1 by 2 to the power n so this can be written as 1 by 2 square plus 1 by 2 plus 1 by 2 to the power 4 and so on.

01:34 So this is equal to 1 by 2 square 1 plus 1 by 2 plus 1 by 2 square and so on.

01:43 So here we see that a is equal to 1 by 4 and r over here is 1 by 2.

01:48 So mod r is less than 1.

01:51 So the series converges.

01:53 And next we find the sum.

01:55 So sum is a by 1 minus r.

01:58 So this on simplification comes out to be equal to half.

02:03 So this is equal to half.

02:05 Next we are given the series as summation n is equal to 0 to infinity, 2 to the power n by 9 to the power 2n plus 1.

02:18 So this becomes equal to 1 by 9 plus 2 by 9 cube plus 4 by 9 to the power 5 and so on.

02:29 So this can be written as 1 by 9 1 plus 2 by 9 square plus 4 by 9 to the power 4 and so on.

02:39 So from here we find a and r.

02:42 So from here a is equal to 1 by 9 and r is equal to 2 by 9 square.

02:48 So mod r here also is less than 1.

02:51 So the series converges and sum is equal to a by 1 minus r.

02:57 So on simplifying, this comes out to be equal to 7 by 79.

03:04 Next we are given part d, submission, n is equal to 5 to infinity, 8 by 9 to the power n.

03:15 So this is equal to 8 by 9 to the power 5 plus 8 by 9 to the power 6 plus 8 by 9 to the power 7 and so on.

03:26 So this comes out to be equal to 8 by 9 to the power 5 common...

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