Infinite Geometric Series

kunny

Find the range of the value for n- th term of the infinite geometric series whose sum is 1.

kunny

Myth

I can't understand statement Blush

**Posted:** Sat Dec 18, 2004 12:38 am

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Myth is out of here

kunny

Sorry for my poor English, Myth.

I have edited. How about this one?

kunny

**Posted:** Sat Dec 18, 2004 7:55 am

Myth

Ok

**Posted:** Sat Dec 18, 2004 7:56 am

_________________
Myth is out of here

Kent Merryfield

Two notational or conventional understandings in my answer:
(1) The index of summation starts with $n=0.$
(2) $1+0+0+0+\cdots$ is considered a geometric series.

With those understandings, the range of allowed values for the $n$ th term is:

For $n=0,$ the interval $(0,2).$
For $n$ even, $n>0$ , the interval $[0,2).$

For $n$ odd, the interval $\left(-2,\frac{n^n}{(n+1)^{n+1}}\right].$

If you reject the convention in (2), then the answers are the same except for the deletion of $0$ from the sets for $n>0.$

If you restrict this to positive term series, then for $n=0$ we get the interval $(0,1)$ and for $n>0$ we get the interval $\left(0,\frac{n^n}{(n+1)^{n+1}}\right].$

**Posted:** Sat Dec 18, 2004 10:14 am

kunny

To:Kent Merryfield

We restrict $n$ to be natural number.

Best regards

kunny

**Posted:** Sat Dec 18, 2004 3:50 pm

Kent Merryfield

**Posted:** Sat Dec 18, 2004 4:03 pm

kunny

Oh! I see. Smile

Here is the solution.

Let the n-th term of geometric sequence be $a_n$ ,the answer is

For $n=1,\ 0<a_n<2$
For $n=2,4,6,\cdots\ -2<a_n\leqq \frac{(n-1)^{n-1}}{n^n}$
For $n=1,3,5,\cdots\ 0\leqq a_n<2$

**Posted:** Sat Dec 18, 2004 5:09 pm