To all of my math people!?

"as n--> infinity, the terms take on the value of 0. Also, the series converges on 1, or n/(n+1) if you take the sum of the first n terms. Really don't know where else to go from here!"

If you've already been able to prove that the sum of the first n terms is n/(n+1) for even n, and (n+2)/(n+1) for odd n, then you've already done most of the work. Remember, the sum of an infinite series is simply the limit of the partial sums as n→∞. So, you simply take the limit of n/(n+1) as n→∞, and the same for (n+2)/(n+1), and if the limit of the odd-numbered terms and the even-numbered terms be the same, they give you the limit of the entire sequence of partial sums and therefore the sum of the series. In this case, both of these limits are fairly obviously 1, so the sum of the series is 1.

Proving that the partial sums have this form is a little harder. Since I want to feel like I'm actually helping you, I'll prove the partial sums for the odd numbered terms are of the form (n+2)/(n+1). First, note that this is true when n=1. We now show that if it is true for some odd n, it is also true for n+2:

[k=1, n+2]∑(-(-1)^k*(2k+1)/(k(k+1)))

[k=1, n]∑(-(-1)^k*(2k+1)/(k(k+1))) - (2n+3)/((n+1)(n+2)) + (2n+5)/((n+2)(n+3))

(n+2)/(n+1) - (2n+3)/((n+1)(n+2)) + (2n+5)/((n+2)(n+3))

((n+2)²(n+3) - (2n+3)(n+3) + (2n+5)(n+1))/((n+1)(n+2)(n+3))

((n³+7n²+16n+12) - (2n²+9n+9) + (2n²+7n+5))/((n+1)(n+2)(n+3))

(n³+7n²+14n+8)/((n+1)(n+2)(n+3))

((n+1)(n+2)(n+4))/((n+1)(n+2)(n+3))

(n+4)/(n+3)

((n+2)+2)/((n+2)+1)

Therefore by induction, all partial sums of odd n have the form (n+2)/(n+1). One may prove the formula for the even terms in a similar fashion.

---- Alright! First to solve this problem, it will be easier if you are taking or if you have taken any Calculus class ------ This problem is a good fun!

CONVERSION:

Now, Here's one way to get the answer: But, you just have to add the numbers by yourself, so, that's not a big deal, is it? Before I bark more, Let's get it on!

Alright! I used:

SUBMISSION from "1" to "n" then the variable that I used for calculation is,

The Variable inside "submission" = ((-1) ^ (n-1)) X Integration from "0" to "n" for [ (n+1) (2n+1) dx]

or,

The Variable inside "submission" = ((-1) ^ (n-1)) X n X (n+1) X (2n+1)

" ^ " = to the power of,

" X " = times

where,

1.) "n" is the number of terms and "n" must be greater than or equal to 1, and

2.) "dx" is the small element that corresponds to the increment in the number of terms.

TRY IT!!!

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

Now, what does this series TELLS US ABOUT ??

1.) When you divide each term by "6", you will get it in the form of the sum of terms as " n ^ 2 ", where " n " is the number of terms.

Again, LET x = 1 (Don't relate it with dx, this is the next "x" term, Say it "x1"..But to make my writing clear, I am using "x" again instead of any other term. )

Also, Let "S" be the sum of terms.

2.) Sum upto the 1st term must be = " 6 * x^2 "

-->Sum upto the 2nd term will be = " 6 * (x+1)^2 ", but it will be negative, right!

--> Sum upto the 3rd term will be = " [[6 * (x+2)^2] + S1], while it will be positive. [ Where, S1 = Sum upto the 1st term]

--> Sum upto the 4th term will be = " [[6 * (x+3)^2 ]+S2] ", while it will be negative again. [ Where, S2 = Sum upto the 2nd term]

-->Sum upto the 5th term will be = " [ 6 * (x+4)^2]+S3] ", while it will be positive again. [ Similarly, S3 = Sum upto the 3rd term]

===> As you can see a pattern over here, you can tell that, for the sum upto the odd numbers, the sum will always be positive, whereas, for the sum upto the even numbers, the sum will always be negative.]

Now, you can also make the form of the sums as:

"S" upto "nth" term = ((-1) ^ (n-1)) * [[6(n^2) + Sum upto the (n-2)th term].

Here comes the fun part baby!

Let's take the "S3" - Sum upto the 3rd term.

Since, S3 = 6 ((3) ^ 2) + S1, so;

=>S3 = 6 ((3) ^ 2) + 6 ((1) ^ 2), Take "6" common, you will see:

=>S3 = 6 [ (3) ^ 2 + (1) ^ 2] i.e. 6 times ( 3 sqaure + 1 square )

Again, let's see for S5,

=>S5 = 6 ((5) ^ 2) + S3,

=> S5 = 6 ((5) ^ 2) + 6 ((3) ^ 2) + 6 ((1) ^ 2) -- [Since, Taking S3 from above]

=> S5 = 6 times ( 5 sqaure + 3 square + 1 square ) -- [Since, Taking "6" as a common factor]

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

Main Theory (for this series): Thus, as the series goes on, you will find that the sum upto the odd number(th) term is going to be:

" 6 times the sum of the sqaure of the odd numbers upto that term "

Similarly, as the series goes on, you will also find that the sum upto the even number(th) term is going to be:

* 6 times the sum of the square of even numbers upto that term." , while it will be negative [ Condition: I have already mentioned it at the top]

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

You can try for even numbers!!

Have fun!!

1 * 2 = 2

[(-1)^(n-1)(2n+1)] / [n(n+1)]

OR

[(-1)^(n+1)(2n+1)] / [n(n+1)]

Looking at the series, you can see that the numerator is the sum of the term index (n) and the next term (n+1), and the denominator is the product of same.

The sign change from term to term can be handled by raising (-1) to a variable power: If the first term needs to be positive then raise (-1) to (n-1) or (n+1). If the first term is negative, then raise (-1) to n.

As for the number that the series converges to: time to pull out tha scratch paper...

1st let us have a look at the nth term ingorning the sign , we shall put it later

it is (n+n+1)/n(n+1) = 1/n + 1/n+ 1

alternate terms are -ve so

1st term = 1 + 1/2

2nd term = - 1/2 - 1/3

the 2nd term of nth term and 1st term of n+1 cancell as values are same and sign opposite

if we add this way the sum upto nth term = 1 + (-1)^(n+1)/n+1

if n is odd we get 1 + 1/n+1 = (n+2)/(n+1)

if n is even 1- 1/(n+1) = n/(n+1)

as n->infinte then whether odd or even it is 1

If I knew the answer I could see if I calculated it right and tell you how but I haven't done fractions in so long. I remember we had to make the fraction into a whole number. Was it 3 times the top number and add the bottom number or somethung like that with that first number? Anyway, sure wish I remembered. It was 30 years ago. I used to love doing fractions.

i'm maths instructor sending you this innovations: a million) do no longer restore unique time, 2) interest is the motivating element to study mathematics. 3) very maximum major element to be said is "practice Examples and routines in elementary words. 4) do no longer bypass over any financial ruin. 5) if you're a pupil of state board,do no longer study some thing from out area the textual content e book. 6) Refer previous twelve months questions. 6) if ISC, you should refer extra books. 7) interest and exertions with sensible might want to convey you 100p.c.successes

I'm not sure that series does converge. If you do an integral test, it doesn't seem to work.

OMg that hella hard sh!t