The ratio test fails when \(\displaystyle \lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | =1\). Otherwise the series converges absolutely if \(\displaystyle \lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | < 1\), and diverges if \(\displaystyle \lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | >1\).

Testing the series \(\displaystyle \sum_{k=0}^{\infty} k\), we have



\(\displaystyle \lim_{k\to\infty} \left |\frac{a_{k+1}}{a_k} \right | = \lim_{k\to\infty} \left |\frac{(k+1)}{(k)} \right | = \lim_{k\to\infty} 1+ \frac{1}{k} = 1.\)

Hence the ratio test fails here. (It is likely obvious to the reader that this series diverges already. However, we must remember that all intuition in mathematics requires rigorous justification. We are attempting that here.)

As required by this question we will have to use the ratio test. \(\displaystyle L=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|\) if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge.

To do so, we will need to compute : \(\displaystyle \frac{a_{n+1}}{a_{n}}\). In our case:

 \(\displaystyle a_{n}=\frac{1}{a_{n}}\)

Therefore

\(\displaystyle \frac{a_{n+1}}{a_{n}}=(\frac{n}{n+1})^{\alpha}\frac{n+1}{n+2}\).

We know that \(\displaystyle \lim_{n\rightarrow \infty }\frac{n}{n+1}=1\)

This means that

\(\displaystyle \frac{a_{n+1}}{a_{n}}=(\frac{n}{n+1})^{\alpha}\frac{n+1}{n+2}=1\forall\quad \alpha\)

Since L=1 by the ratio test, we can't conclude about the convergence of the series.

To be able to use to conclude using the ratio test, we will need to first compute the ratio then use \(\displaystyle L=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|\) if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge. Computing the ratio we get,

\(\displaystyle \frac{a_{n+1}}{a_{n}}\).

We have then:

\(\displaystyle a_{n+1}=\frac{1}{72(n+1)+89}=\frac{1}{72n+161}\)

Therefore have :

 \(\displaystyle \frac{a_{n+1}}{a_{n}}=\)\(\displaystyle \frac{72n+89}{72n+161}\) 

It is clear that \(\displaystyle \lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_{n}}=1\).

By the ratio test , we can't conclude about the nature of the series.

Let \(\displaystyle a_{n}\) be the general term of the series. We will use the ratio test to check the convergence of the series. 

\(\displaystyle L=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|\) if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge.

We need to evaluate,

\(\displaystyle \frac{a_{n+1}}{a_{n}}\) we have:

\(\displaystyle a_{n+1}= \frac{n+1}{(n+1)^\alpha +1},a_{n}=\frac{n}{n^\alpha +1}\).

Therefore:

\(\displaystyle \frac{a_{n+1}}{a_{n}}=\frac{n^\alpha +1}{(n+1)^\alpha +1}\frac{n+1}{n}\). We know that,

\(\displaystyle \lim_{n\rightarrow \infty}\frac{n}{n+1}=1\) and therefore

\(\displaystyle \lim_{n\rightarrow \infty}\frac{n^\alpha +1}{(n+1)^\alpha +1}=1\)

This means that :

\(\displaystyle \lim_{n\rightarrow \infty}\frac{a_{n}}{a_{n+1}}=1\).

By the ratio test we can't conclude about the nature of the series. We will have to use another test.