The discussion revolves around the convergence of the series \(\sum_{n=1}^{\infty} \frac{(-2)^n}{n^n}\), with participants exploring the application of the ratio test to determine its behavior.
The discussion includes various attempts to analyze the series using the ratio test, with some participants suggesting that the series of positive terms converges. There is no explicit consensus, but several lines of reasoning are being explored regarding the convergence of the series.
Some participants note potential missing information or assumptions in the problem setup, particularly regarding the handling of terms in the limit process.
Aerospace93 said:Homework Statement
\sum ftom n=1 to \infty (-2)n/nn.
The Attempt at a Solution
limn->\infty | (-2)n+1/(n+1)n+1) x nn/(-2)n | = |-2|limn->\infty |(n/n+1)n*(1/n+1) |
If it were only (n/n+1) then would the answer be 2e? Either way, how do you sole this the way it is?
Aerospace93 said:Homework Statement
\sum ftom n=1 to \infty (-2)n/nn.
The Attempt at a Solution
limn->\infty | (-2)n+1/(n+1)n+1) x nn/(-2)n | = |-2|limn->\infty |(n/n+1)n*(1/n+1) |
If it were only (n/n+1) then would the answer be 2e? Either way, how do you sole this the way it is?
Zondrina said:Notice that by the ratio test, the series of positive terms will converge and so Ʃan will be absolutely convergent which implies Ʃan will converge.
You can always divide numerator and denominator by n, if you do it consistently (as n!=0).Aerospace93 said:The only thing I am not sure about doing in this problem really is how to deal with the numbers when i divide them by n when I'm working out the limit.