Question

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

Answer 1

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

is divergent.

Explanation:

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.

Answer 2

The series diverges

Explanation:

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.