help with a series: sigma sum w/ infty on top, n=2 on bottom

[Opus89]

I need to solve a particular series of : the sigma sum symbol with infinity on top and n = 2 on the bottom of -1/2^n. I identified this as a geometric sequence with a = -1 and r = 1/2. I only need to answer whether it converges or diverges and I have concluded that it converges since 0<|1/2|<1. Is this reasoning correct?

 

[galactus]

Re: help with a series

Yes, it does converge.

\(\displaystyle \frac{\frac{-1}{4}}{1-\frac{1}{2}}=\frac{-1}{2}\)

 

[pka]

Re: help with a series

I need to solve a particular series of : the sigma sum symbol with infinity on top and n = 2 on the bottom of -1/2^n. I identified this as a geometric sequence with a = -1 and r = 1/2. I only need to answer whether it converges or diverges and I have concluded that it converges since 0<|1/2|<1. Is this reasoning correct?

It depends on which you mean: \(\displaystyle \sum\limits_{n = 2}^\infty {\frac{{ - 1}}{{2^n }}} \text{ OR }\sum\limits_{n = 2}^\infty {\left( {\frac{{ - 1}}{2}} \right)^n } ?\)

You wrote the first one. The sum would be \(\displaystyle \frac{a}{1-r}\).
But you got r correct but missed the first term: \(\displaystyle a=\frac{-1}{2^2}\).

 

[Opus89]

Re: help with a series

I need to solve a particular series of : the sigma sum symbol with infinity on top and n = 2 on the bottom of -1/2^n. I identified this as a geometric sequence with a = -1 and r = 1/2. I only need to answer whether it converges or diverges and I have concluded that it converges since 0<|1/2|<1. Is this reasoning correct?

It depends on which you mean: \(\displaystyle \sum\limits_{n = 2}^\infty {\frac{{ - 1}}{{2^n }}} \text{ OR }\sum\limits_{n = 2}^\infty {\left( {\frac{{ - 1}}{2}} \right)^n } ?\)

You wrote the first one. The sum would be \(\displaystyle \frac{a}{1-r}\).
But you got r correct but missed the first term: \(\displaystyle a=\frac{-1}{2^2}\).

Yea, I meant the first one you listed.