Difficulty in finding interval of convergence with power series I have the following power series:
$$\sum_{n=1}^\infty \frac{(4x+1)^{n}}{n}  $$
When finding the interval of convergence, I am left with the following inequality:
$$ |4x+1|\lt1 $$
How do I go about finding the values of $x$ for which this series converges absolutely if I end up with 0 on the right side?
Thank you!

Another approach, differentiating our series term-by-term we find:
$$\frac{\partial }{\partial x}\sum_{n=1}^\infty \frac{(4x+1)^{n}}{n} = \sum_{n=1}^\infty 4\frac{n(4x+1)^{n-1}}{n} = 4\sum_{n=0}^\infty (4x+1)^{n}$$
This is the geometric series for $\cases{\frac{1}{1-r}\\ r = 4x+1}$
which converges on $|r|<1 \Leftrightarrow |4x+1|<1$ as both you and the other answerer got.
