How do you know if the series #sum 1/(n^(1+1/n))# converges or diverges for (n=1 , ∞) ?

Using the properties of exponents:

#1/n^(1+1/n) = 1/(n * n^(1/n)) = 1/n n^(-1/n)#

Note now that:

#n^(-1/n) = (e^(ln n))^(-1/n) = e^(-ln n /n)#

As:

#lim_(n->oo) ln/n = 0#

and the exponential function #e^x# is continuous, then:

#lim_(n->oo)e^(-ln n /n) = e^((lim_(n->oo) -ln/n )) = e^0 = 1#

Consider now the harmonic series

#sum_(n=1)^oo 1/n#

that we know to be divergent.

Using the limit comparison test:

#lim_(n->oo) (1/n^(1+1/n))/(1/n) = lim_(n->oo) (1/n n^(-1/n))/(1/n) = lim_(n->oo) n^(-1/n) = 1#

we can see that, as the limit of the ratio is finite, the two series have the same character and also:

#sum_(n=1)^oo 1/n^(1+1/n)#

is divergent.

To test the convergence of the series #sum_{n=1}^oo a_n#, where #a_n=1/n^(1+1/n)# we carry out the limit comparison test with another series #sum_{n=1}^oo b_n#, where #b_n=1/n#,

We need to calculate the limit

#L = lim_{n to oo }a_n/b_n = lim_{n to oo} n^{-1/n}#

Now,

#ln L = lim_{n to oo}( -1/n ln n) = 0 implies L=1#

According to the limit comparison test , since this limit is a finite nonzero number, the series #sum_{n=1}^oo a_n# if and only if #sum_{n=1}^oo b_n# converges.

However, it is well known that #sum_{n=1}^oo b_n# diverges, and hence our series diverges.