Rearranging Series

1. andrey21

466
As we know the sum of the series ((-1)^(n-1))/(n) is equal to ln2: I have bin asked to rearrange the series so its sum is SQRT2

Also write out the first 20 terms

How should I go about solving. Set original series equal to SQRT 2 and solve from there? Use trial and error until I arrive at the correct series.

2. micromass

18,679
Staff Emeritus
I'll give a small sketch. I'll leave the details to you.

First take some positive terms such that the sum of these terms is just above sqrt(2).
Then take some negative terms such that the sum of all the terms comes back below sqrt(2).
Then take enough positive terms such that the sum comes above sqrt(2) again.
Continue with this process...

3. andrey21

466
Thanks micromass Im actually gonna come back to that question. While I have u here I have bin posed the following:

Describe the following series and if possible find its limits:

1/4 +1/10+1/18+1/28.....

any ideas??

4. micromass

18,679
Staff Emeritus
Well, let's take a look at the sequence 4,10,18,28,...
What is the fifth element of this sequence?
Can you find a general rule for how to obtain the next element in this sequence?
And: more important, can you find a general term for this sequence?

5. andrey21

466
Well from observing the sequence the next term is 1/40. Correct??

6. micromass

18,679
Staff Emeritus
Yes, so the denumerators are the following:

$$4,4+6,4+6+8,4+6+8+10,4+6+8+10+12,...$$

Now try to find a general rule for this sequence.

HINT: you've probably seen that

$$\sum_{k=0}^n{k}=\frac{n(n+1)}{2}$$

use this...

7. andrey21

466
Well as a rule for the sequence isnt it similar to fibonacci. SO:

fn = fn-1+fn-2

for n>1

8. micromass

18,679
Staff Emeritus
No, this isn't like fibonacci at all. What is the general rule for this sequence?

9. andrey21

466
the next term is just the previous terms added to the new term.

10. micromass

18,679
Staff Emeritus
Indeed. So, if you're given the nth term $$x_n$$. How do you obtain the next term. In other words, fill in the blanks:

$$x_{n+1}=x_n+...$$

HINT:
x_0=4
x_1=x_0+6
x_2=x_1+8
x_3=x_2+10

11. andrey21

466
I think I have the general rule is it:

1/(n+2)^2 +2

so sub in 0

1/(0+2)^2 +0

1/4

sub in 1

1/(1+2)^2 + 1

1/10....

12. micromass

18,679
Staff Emeritus
Yes, that is the general rule. So now you have the series

$$\sum_{n=0}^{+\infty}{\frac{1}{(n+2)^2+n}}$$

13. andrey21

466
Great it says find its limit? Is this the sum?

14. micromass

18,679
Staff Emeritus
Well, to find the limit of a series is not always possible. But in this case it is.
Try to write the series as a telescopic sum. To do that, split up the fraction

$$\frac{1}{(n+2)^2+n}$$

in partial fractions. This will help a great deal...

15. andrey21

466
Oh ok should it look like:

1 = A/(n+2)^2 +B/(n+2) + C/n
Then find values for A B and C?

16. andrey21

466
Giving values of

A = -1/4 B=/1/2 c=1/4

17. micromass

18,679
Staff Emeritus
Nonono. First you need to simplify $$(n+2)^2+n$$. Then you need to factor it. Only then can you begin calculating the partial fractions...

18. andrey21

466
Ah ok so that gives n^2 +5n +4 which factors to:

(n+1)(n+4)

Then solving will give values for A and B. Then wot shall i do???

19. micromass

18,679
Staff Emeritus
Yes, this is correct. So what are your values for A and B?
What is

$$\frac{1}{(n+1)(n+4)}=...$$?

20. andrey21

466
A = 1/3
b= -1/3

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