Difficulty in finding interval of convergence with power series

Question

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!

Maybe it's derivative can help find something. – mathreadler Apr 17 '17 at 19:59 — mathreadler, Apr 17 '17 at 19:59
Note that $$|4x+1|<1\implies -\frac12 <x < 0$$ – Mark Viola Apr 17 '17 at 20:01 — Mark Viola, Apr 17 '17 at 20:01

score 3 · Accepted Answer · answered Apr 17 '17 at 20:05

3

Note that

$$|4x+1|<1\implies -\frac12 <x < 0$$

The series converges absolutely in this interval.

Now, to test the endpoints, we see that for $x=-1/2$, $4x+1=-1$ and the alternating harmonic series converges.

On the other hand, for $x=0$ we see that $4x+1=1$ and the harmonic series diverges.

Therefore, the series $\sum_{n=1}^\infty\frac{(4x+1)^n}{n}$ converges for $-1/2\le x<0$ and diverges elsewhere.

answered Apr 17 '17 at 20:05

Mark Viola

score 1 · Answer 2 · answered Apr 17 '17 at 20:14

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.

2 Answers2