# Using the ratio test to figure if the series is convergent

1. Mar 8, 2013

### Aerospace93

1. The problem statement, all variables and given/known data
$\sum$ ftom n=1 to $\infty$ (-2)n/nn.

3. 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?

2. Mar 8, 2013

### Staff: Mentor

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

3. Mar 8, 2013

### 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?

4. Mar 8, 2013

### Ray Vickson

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

5. Mar 8, 2013

### Zondrina

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.

6. Mar 9, 2013

### Aerospace93

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 im 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.

7. Mar 9, 2013

### Staff: Mentor

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

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