Question

How do you know if the series \[\sum {\dfrac{1}{{{n^{1 + \dfrac{1}{n}}}}}} \] converges or diverges for \[\left( {n = 1,\infty } \right)\] ?

Answer 1

Hint: To test the convergent of the given series \[\sum\limits_{n = 1}^\infty {\dfrac{1}{{{n^{1 + \dfrac{1}{n}}}}}} \] , we need to apply limit comparison test with another series i.e., \[\sum\limits_{n = 1}^\infty {\dfrac{1}{n}} \] . Hence, by using the limits we can find, whether the given series converges or diverges for \[\left( {n = 1,\infty } \right)\] .

Complete step-by-step answer:
 Let us write the given series:
 \[\sum {\dfrac{1}{{{n^{1 + \dfrac{1}{n}}}}}} \]
To test the convergence of the given series, we need to apply a limit comparison test with another series. Hence, let us consider the harmonic series \[\sum\limits_{n = 1}^\infty {\dfrac{1}{n}} \] , to know the divergence.
Let,
 \[{a_n} = \dfrac{1}{{{n^{1 + \dfrac{1}{n}}}}}\] and \[{b_n} = \dfrac{1}{n}\]
Hence, by using the limit comparison test, we can find the series converges or diverges i.e., we need to calculate the limit:
 \[L = \mathop {\lim }\limits_{n \to \infty } \dfrac{{{a_n}}}{{{b_n}}}\]
Substitute the value of \[{a_n}\] and \[{b_n}\] as:
 \[L = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\dfrac{1}{{{n^{1 + \dfrac{1}{n}}}}}}}{{\dfrac{1}{n}}}\]
  \[L = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\dfrac{1}{n}{n^{ - \dfrac{1}{n}}}}}{{\dfrac{1}{n}}}\]
Now, simplifying the terms with respect to n we get:
 \[L = \mathop {\lim }\limits_{n \to \infty } {n^{ - \dfrac{1}{n}}}\]
Now,
 \[\ln L = \mathop {\lim }\limits_{n \to \infty } \left( { - \dfrac{1}{n}\ln n} \right)\]
As, \[n \to \infty \] we get:
 \[ \Rightarrow \ln L = 0\]
 \[ \Rightarrow L = 1\]
According to the limit comparison test, since this limit is a finite non-zero number, the series \[\sum\limits_{n = 1}^\infty {\dfrac{1}{{{n^{1 + \dfrac{1}{n}}}}}} \] converges if and only if \[\sum\limits_{n = 1}^\infty {\dfrac{1}{n}} \] converges.
As the limit of the ratio is finite, the two series have the same character and also \[\sum\limits_{n = 1}^\infty {\dfrac{1}{{{n^{1 + \dfrac{1}{n}}}}}} \] is divergent. However, it is well known that \[\sum\limits_{n = 1}^\infty {\dfrac{1}{n}} \] diverges, and hence the given series diverges.

Note: The key point is that, as the limit of the ratio is finite, the two series have the same character and also:
 \[\sum\limits_{n = 1}^\infty {\dfrac{1}{{{n^{1 + \dfrac{1}{n}}}}}} \] is divergent. We must also note that if the limit \[\dfrac{{{a_n}}}{{{b_n}}}\] is positive, then the sum of \[{a_n}\] converges if and only if the sum of \[{b_n}\] converges and if the limit of \[\dfrac{{{a_n}}}{{{b_n}}}\] is zero, and the sum of \[{b_n}\] converges, then the sum of numerator term \[{a_n}\] also converges the series.