Question

How do you test the series `Sigma n^-n` from n is `[1,oo)` for convergence?

Answer 1

Using the ratio test

Explanation:

The ratio test finds the ratio between terms `u_k and u_(k+1)` as `k` tends to infinity, and if `abs(u_(k+1)/u_k)<1 (k->oo)`, then the series is found to be convergent, as it means that terms are getting progressively smaller and tending towards 0.

Hence, in order to test `sum n^-n, n in [1,oo)` for convergence, we find the ratio `abs(u_(k+1)/u_k)(k->oo)`:

`k->oo abs(u_(k+1)/u_k)`
`= k->ooabs((k+1)^-(k+1)/k^-k)`
`=k->oo abs(k^k/(k+1)^(k+1))`
`=k->oo abs(k^k/(k+1)^k*1/(k+1))`
`=k->oo abs((k/(k+1))^k*1/(k+1))`
`=k->oo abs((1/(1+1/k))^k*1/(k+1))`
`because k->oo 1/k=0`
`therefore k->ooabs((1/(1+1/k))^k)=k->ooabs((1/(1+0))^k)=abs(1)=1`
`therefore k->ooabs((1/(1+1/k))^k*1/(k+1))=k->ooabs(1*1/(k+1))`

`=k->ooabs(1/(k+1))=0`

`because k->oo abs(u_(k+1)/u_k)=0`
i.e. `k->oo abs(u_(k+1)/u_k)<1`
`therefore` By the ratio test, the series `sum n^-n` converges from `n in [1,oo)`