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.