Rearranging Series to Equal SQRT2: How to Solve

[andrey21]

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.

 

[micromass]

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...

 

[andrey21]

Thanks micromass I am actually going to 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??

 

[micromass]

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?

 

[andrey21]

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

 

[micromass]

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...

 

[andrey21]

Well as a rule for the sequence isn't it similar to fibonacci. SO:

fn = fn-1+fn-2

for n>1

 

[micromass]

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

 

[andrey21]

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

 

[micromass]

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

 

[andrey21]

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...

 

[micromass]

Yes, that is the general rule. So now you have the series

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

 

[andrey21]

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

 

[micromass]

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...

 

[andrey21]

Oh ok should it look like:

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

 

[andrey21]

Giving values of

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

 

[micromass]

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...

 

[andrey21]

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?

 

[micromass]

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

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

 

[andrey21]

A = 1/3
b= -1/3

 

[micromass]

Yes, so we have the series

$$\sum_{n=1}^{+\infty}{\frac{1}{3(n+1)}-\frac{1}{3(4+n)} }$$

Now you have to use a little trick. Try to write out this series for the first 10 terms (without adding any of the fractions). Do you see something cool?

 

[andrey21]

Alot of the terms cancel out and I am left with:

1/3+ 1/6+1/9 -1/36-1/39-1/42

for first 10 terms:

 

[micromass]

Yes, that's exactly what we wanted!

Can you do this in general now? If you calculate the first k terms, there will be a lot of terms cancelling out. So which of the terms remain?

If you don't see this immediately, try taking k some other values. Try taking k=20 and k=30. Then you will be ready to handle the general situation...

 

[andrey21]

Well in this case it is the first 3 and last 3 terms which remain correct?

 

[micromass]

Indeed. So you've shown that

$$\sum_{n=1}^{k}{\left(\frac{1}{3(n+1)}-\frac{1}{3(n+4)}\right)}=\frac{1}{3}+\frac{1}{6}+\frac{1}{9}-\frac{1}{3(k+2)}-\frac{1}{3(k+3)}-\frac{1}{3(k+4)}$$

Now you can easily take the limit $$k\rightarrow +\infty$$

 

[andrey21]

so as k tends to infinty the limit is 11/18.

 

[micromass]

Correct! It seems like you've got it!

 

[andrey21]

Fantastic thanks micromass. I do have one final question very similar.

given the sequence:

1+1/9+1/25+1/49...

I know this can be written as:

sum 1/(2n+1)^2

correct??

What does this converge to??

 

[micromass]

Do you know what the sum of

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

is?? If you don't know the above sum, then your question is very difficult...

 

[andrey21]

Yes the sum of 1/n^2 is pi^2/6 correct??

 

[micromass]

Ah, yes. this is good.
Now, you've got that

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

You know two of the above series...

 

[andrey21]

ah yes i see simply solve to get:

1/4n^2 which equals pi^2/24 correct??

Then subtract to leave me value for 1/(2n+1)^2

 

[micromass]

Correct!

 

[andrey21]

Great thanks micromass :)

 

[andrey21]

I do have another series question micromass:
I have bin given the series:

5/4 + 1 + 4/5 + 16/25+...

It says describe what type of series this is?

Shall i try find general formula again?