Using the ratio test to figure if the series is convergent

[Aerospace93]

Homework Statement

\sum ftom n=1 to \infty (-2)ⁿ/nⁿ.

The Attempt at a Solution

lim_n->\infty | (-2)ⁿ⁺¹/(n+1)ⁿ⁺¹⁾ x nⁿ/(-2)ⁿ | = |-2|lim_n->\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?

 

[mfb]

I think there are brackets missing.
(n/(n+1)^n) is smaller than 1, that is sufficient here. The other factor is more interesting.

 

[fortissimo]

Another way to look at it: The growth rate of n^n is faster than that of n!. If you replace n^n with n!, where does that leave you at?

 

[Ray Vickson]

Homework Statement

\sum ftom n=1 to \infty (-2)ⁿ/nⁿ.

The Attempt at a Solution

lim_n->\infty | (-2)ⁿ⁺¹/(n+1)ⁿ⁺¹⁾ x nⁿ/(-2)ⁿ | = |-2|lim_n->\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?

If n > 3 then (2/n)^n < (2/3)^n.

 

[STEMucator]

Homework Statement

\sum ftom n=1 to \infty (-2)ⁿ/nⁿ.

The Attempt at a Solution

lim_n->\infty | (-2)ⁿ⁺¹/(n+1)ⁿ⁺¹⁾ x nⁿ/(-2)ⁿ | = |-2|lim_n->\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?

Notice that by the ratio test, the series of positive terms will converge and so Ʃa_n will be absolutely convergent which implies Ʃa_n will converge.

 

[Aerospace93]

Notice that by the ratio test, the series of positive terms will converge and so Ʃa_n will be absolutely convergent which implies Ʃa_n will converge.

By the ratio test I'm left with: 2 x limx->\infty |nⁿ/(n+1)ⁿ⁺¹⁾| which is the same as what i previously wrote: 2 x limx->\infty [n/(n+1)]ⁿ*1/(n+1).

Dividing by the largest power of the polynomial in the denominator:
2 x limx->\infty [1/(1+1/n)]ⁿ*1/(n+1).-
so it becomes: 2 x limx->\infty |1ⁿ*[1/(1+n)] = 2 x limx->\infty 1ⁿ*0=0
ratio test = 0>1 therefore the series is absolutely convergent and therefore convergent.

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.

 

[mfb]

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.

You can always divide numerator and denominator by n, if you do it consistently (as n!=0).