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## NN321 3 years ago how do you find the sum from n=1 to infinity of n(1/4)^(n-1)

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1. perl

did you try using the ratio test

2. NN321

I thought the ratio test tells you if it is convergent or divergent not what the actual sum is. By the ratio test it is convergent.

3. perl

|dw:1353738999087:dw|

4. perl

is that the question>

5. NN321

yes

6. perl

the expression inside looks close to the derivative of 1/4 ^n-1

7. perl

the answer is 16/9

8. NN321

how did you get that?

9. perl

lets look at the partial sums

10. perl

take the sum from n=1 to m

11. cinar

16/9

12. perl

do you have a justification

13. cinar

|dw:1353747498242:dw|

14. cinar

derivative

15. cinar

x=1/4

16. cinar

|dw:1353747595344:dw|

17. perl

what about the limits ?

18. cinar

not needed

19. perl

i could start the second limit at n = 2

20. NN321

how did you get 1/(1-x)^2 from the step above?

21. cinar

|dw:1353747789535:dw|

22. cinar

take the derivative both sides

23. NN321

ok. cool. thank you

24. cinar

(:

25. cinar

you can also take integral sometimes..

26. perl

that formula is incorrect

27. perl

|dw:1353740583496:dw|

28. perl

|dw:1353740607290:dw|

29. perl

very sloppy indeed

30. cinar
31. cinar

I did not notice your series start from 1, answer will be (16/9)-1

32. cinar

|dw:1353748518528:dw|

33. perl

the answer is correct , it is 16/9

34. perl

|dw:1353741295190:dw|

35. perl

the first term is zero, so we can ignore it

36. NN321

the original solution was correct. the left hand side is defined at n=1, the RHS doesn't have to be defined at 1 because its the formula for the sum of geometric series i.e. r=x.

37. perl

|dw:1353741624657:dw|

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