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

Question

Evelyn Agard · Accepted Answer

undefined To test the series $ \sum_{n=1}^{\infty} n^{-n} $ for convergence, you can use the Ratio Test.

The Ratio Test states that if $ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L $, then:

- If $ L < 1 $, the series converges absolutely.
- If $ L > 1 $, the series diverges.
- If $ L = 1 $, the test is inconclusive.

For the given series $ \sum_{n=1}^{\infty} n^{-n} $, let's find the ratio $ \frac{a_{n+1}}{a_n} $:

$$ \frac{a_{n+1}}{a_n} = \frac{(n+1)^{-(n+1)}}{n^{-n}} = \frac{\frac{1}{(n+1)^{n+1}}}{\frac{1}{n^n}} = \frac{n^n}{(n+1)^{n+1}} $$

Taking the limit as $ n \to \infty $:

$$ \lim_{n \to \infty} \frac{n^n}{(n+1)^{n+1}} = \lim_{n \to \infty} \frac{n^n}{n^{n+1}} = \lim_{n \to \infty} \frac{1}{n} = 0 $$

Since $ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 < 1 $, by the Ratio Test, the series $ \sum_{n=1}^{\infty} n^{-n} $ converges absolutely.

William Abdalla · Answer

Using the ratio test

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)#