# convergent or divergent, find its sum

• Nov 24th 2007, 06:06 AM
keith
convergent or divergent, find its sum
Here is the problem:
geometric series is: n=1 to infinity e^n/2^n
here is what I did: (e/2)^n r= e/2 e/2>1
since r>1 the series diverges, is this correct? Do I need to explain this more for a complete answer?

The other one is a geometric series n=0 to infinity 1/Sqrt(2)^n
I plugged in 0 for n and my series is as follows:
1 +1/Sqrt(2) + 1/2 + 1/sqrt(2)^3 + ...
this series diverges, do I need to explain these further?
Am I doing these correctly?
Thank you,
Keith
• Nov 24th 2007, 06:29 AM
TKHunny
I'll buy the first one. You demonstrated that the individual terms do not approach zero. That is sufficient for divergence.

For the second, all you have shown is the first few terms. You made no conclusions about them. Do they approach zero? Does it look like any familiar series, maybe a Geometric Series?
• Nov 24th 2007, 07:11 AM
keith
Quote:

Originally Posted by TKHunny
I'll buy the first one. You demonstrated that the individual terms do not approach zero. That is sufficient for divergence.

For the second, all you have shown is the first few terms. You made no conclusions about them. Do they approach zero? Does it look like any familiar series, maybe a Geometric Series?

I think I understand what you are saying, so showing more entries in the series we can see it approaches zero. I went to 1/((sqrt(2)^15)) an it was .0055242717, which is going closer and closer to zero, therefore the series diverges.
Is there a test that I am overlooking to prove this? Am I going in the right direction?
Thank you,
Keith
• Nov 24th 2007, 08:20 AM
Plato
$\displaystyle \frac{1}{{\left( {\sqrt 2 } \right)^n }} = \left( {\frac{1}{{\sqrt 2 }}} \right)^n \;\& \;\frac{1}{{\sqrt 2 }} < 1$.
Is that right?
• Nov 24th 2007, 08:51 AM
keith
convergent or divergent, find its sum
If I think about this like the area under the curve, where the area is infinite, so the sum of this series must be infinite and the series divergent.
Thank you,
Keith
• Nov 24th 2007, 09:08 AM
Plato
Quote:

Originally Posted by keith
If I think about this like the area under the curve, where the area is infinite, so the sum of this series must be infinite and the series divergent. Is this a good way of thinking about this?

Absolutely not.
How much do you know about geometric series?
What is the main idea about the common ratio?
• Nov 24th 2007, 09:36 AM
keith
convergent or divergent, find its sum
The common ratio is the sum of a geometric series is the factor, I will say for example n=1 to infinity 1/2^n where a= first term which is 1/2 and r= 1/2
r= commom ratio and this series: 1/2+1/4+1/8+1/16 +...+1/2^n +... =1
so, this sum seems to be 1

so the sum of 1/(sqrt(2)^n) looks like it is approaching zero
I guess I just don't get it
Thank you,
Keith
• Nov 24th 2007, 09:46 AM
Plato
For any integer J the geometric series sums as follows.
$\displaystyle \sum\limits_{n = J}^\infty {a\left( r \right)^n } = \frac{{ar^J }}{{1 - r}}\;\mbox{if}\;\left| r \right| < 1.$

Please note: $\displaystyle \color{red}\;\mbox{if}\;\left| r \right| < 1.$