Find the sum of the series, if it converges. Otherwise, enter DNE. Determine whe

Question

Find the sum of the series, if it converges. Otherwise, enter DNE.

Determine whether the geometric series is convergent or divergent.

8 + 7 + 49/8 + 343/64 +......

If it is convergent, find its sum.

find the sum

(b) Find the sum of the series for those values of x.

Find the sum of the series, if it converges. Otherwise, enter DNE. Sum n=1 to infinity ((1/n-1/(n+1))) Determine whether the geometric series is convergent or divergent. 8 + 7 + 49/8 + 343/64 +...... If it is convergent, find its sum. Consider the following series. Sum n=1 to infinity [(0.8)**(n-1)-(0.7)**n] find the sum Consider the following series. Sum n=1 to infinity x**n/(9**n) (a) Find the values of x for which the series converges. ( , ) (b) Find the sum of the series for those values of x.

Explanation / Answer

1/n - 1/(n+1)

Lets write out first few terms :

1/1 - 1/2 + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + .... (1/inf + 1/inf + 1)

Everything cancels except first and last terms :

1 + 1/(inf + 1)

1 + 1/inf

1 + 0

1 ----> ANSWER

So, it converges

------------------------------------------------------------------------------

8 + 7 + 49/8 + 343/64 + ....

Common ratio = 7/8

And since this common ratio, r satisfies -1 < r < 1, this series CONVERGES

Sum = a1 / (1 - r)

a1 --> first term = 8
r --> common ratio = 7/8

Sum = 8 / (1 - 7/8)

Sum = 8 / (1/8)

Sum = 64 and it converges ----> ANSWER

-------------------------------------------------------------------------------

The next one can be split into two series

First :
(sigma 1 to inf) (0.8)^(n - 1)
First term, a1 = 0.8^(1-1) --> a1 = 1
Common ratio = 0.8
Sum of first series = 1 / (1 - 0.8) ---> 1 / 0.2 ---> 5

Second :
(sigma from 1 to inf) (0.7)^n
First term, a1 = (0.7)^1 --> 0.7
Common ratio = 0.7
Sum of the second series = 0.7 / (1 - 0.7) --> 0.7 / 0.3 --> 7/3

So, the total sum is :

5 - 7/3

15/3 - 7/3

8/3 ---> ANSWER

---------------------------------------------------------------------------------

The term is (x/9)^n, a geometric series with common ratio = x/9
So, the rule for convergence is t hat the r must lie within -1 < r < 1

So, -1 < x/9 < 1

-9 < x < 9

So, (-9 , 9) ----> ANSWER for a

Sum = a1 / (1 - r)

First term, a1 = (x/9)^1 --> x/9
Common ratio = x/9

Sum = (x/9) / (1 - x/9)

Sum = (x/9) / (9 - x)/9

Sum = x / (9 - x) ---> ANSWER for b

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.