It is important to distinguish between sum_{n=1}^{infty} n^{b} quad text { and } quad sum_{n=1}^{infty} b^{n} What name is given to the first series? To the second? For what values of b does the first series converge? For what values of b does the second series converge?

Given:
The series $\sum _{n=1}^{\mathrm{\infty }}{n}^b\text{ and }\sum _{n=1}^{\mathrm{\infty }}{b}^n$
As,
$\sum _{n=1}^{\mathrm{\infty }}{n}^b=1^b+2^b+3^b+4^b+...$
Thus it is an arithmetic increasing power series.
Similarly,
$\sum _{n=1}^{\mathrm{\infty }}{b}^n=b+b^2+b^3+b^4+...$
Here first term is b and common ratio is also b.
Thus it is a geometric power series.
Ratio test:
Consider the series: $\sum _{n=1}^{\mathrm{\infty }}{a}^n$
Let $L=\underset{n\to \mathrm{\infty }}{lim}|\frac{a_{n+1}}{a_n}|$
If L< 1, then series is convergent.
If L>1, then series is divergent and
If L=1, then test is inconclusive.
For $\sum _{n=1}^{\mathrm{\infty }}{n}^b$ the nth term $a_n=n^b$
Thus $L=\underset{n\to \mathrm{\infty }}{lim}|\frac{(n+1{)}^b}{n^b}|$
$=\underset{n\to \mathrm{\infty }}{lim}|(\frac{n+1}{n}{)}^b|$
$=\underset{n\to \mathrm{\infty }}{lim}|(1+\frac{1}{n}{)}^b|$
$=(1+0{)}^b$
$=1$
As L=1, test is inconclusive.
Series is convergent if $L<1$.
But for any value of b it will never be less than 1.
Therefore, for any value of b, the series $\sum _{n=1}^{\mathrm{\infty }}{n}^b$ is not convergent.
For $\sum _{n=1}^{\mathrm{\infty }}{b}^n$ the nth term of series is $a_n=b^n$
Thus
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{b^{n+1}}{b^n}|$
$=\underset{n\to \mathrm{\infty }}{lim}|b|$
$=b$
Series is convergent if $L<1$.
Thus $b<1$.
Therefore, for $b<1$, the series $\sum _{n=1}^{\mathrm{\infty }}{b}^n$ is convergent.