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

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.

Is this a good way of thinking about this?

Thank you,

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