Can I use the index of a series for help with divergence? I was studying this series:
$$\sum_{n=2}^{\infty}\dfrac{5}{7n+28}$$
I know that it's an increasing, monotone sequence.
Also, I know I can rewrite as:
$$\sum_{n=2}^{\infty}\dfrac{5}{7(n+4)} = \dfrac{5}{7} \cdot \sum_{n=2}^{\infty}\dfrac{1}{n+4}$$
Also, I know that the following series is called the Harmonic Series and can be shown to diverge by over-estimating grouped terms (essentially at a point it is worse than adding $1/2$ over and over).  (Also, by Cauchy Condensation):
$$\sum_{n=1}^{\infty}\dfrac{1}{n}$$
I tried to setup a comparison test, but I don't think it works since the Harmonic is always larger:
$$0 \lt \dfrac{1}{n+4} \lt \dfrac{1}{n}$$
Is it legitimate to massage the index without affecting divergence/convergence?
For example,
The first few terms in the sequence of the series:
$$\sum_{n=2}^{\infty}\dfrac{1}{n+4}$$
Are $$\dfrac{1}{6},\dfrac{1}{7},\dfrac{1}{8},\dfrac{1}{9},...$$
Which looks just like the Harmonic just starting off at a different place.
So can I somehow change
$$\sum_{n=2}^{\infty}\dfrac{1}{n+4}$$  to $$\sum_{n=-3}^{\infty}\dfrac{1}{n+4}$$
to show it diverges?  Or am I just way off track here?
 A: Notice that by changing the index we have
$$\sum_{n=2}^\infty\frac1{n+4}=\sum_{n=6}^\infty\frac1n$$
so the series is divergent. Notice also that the nature of a series doesn't depend on the first few terms which means that the two series $\sum\limits_{n\ge1}u_n$ and $\sum\limits_{n\ge n_0}u_n$ (for any $n_0$) have the same nature (both convergent or both divergent).
A: I'd encourage playing with the algebra a little bit more if you want to make the harmonic series pop out. Starting all the way back at  $$\dfrac{5}{7} \cdot \sum_{n=2}^{\infty}\dfrac{1}{n+4}$$ then  $$\dfrac{5}{7} \cdot \sum_{n=2}^{\infty}\dfrac{1}{n+4} = \dfrac{5}{7} \cdot \sum_{n=1}^{\infty}\dfrac{1}{n+5} \\ \geq \dfrac{5}{7}\sum_{n=1}^{\infty}\dfrac{1}{(6n-5)+5} \\ =  \dfrac{5}{7}\sum_{n=1}^{\infty}\dfrac{1}{6n} \\ = \dfrac{1}{7}\sum_{n=1}^{\infty}\dfrac{1}{n}$$ On the line where I introduced the "$\geq$", I know the relationship holds because $$\frac{1}{n+5} \geq \frac{1}{(6n-5)+5} = \frac{1}{6n} \quad \text{for all} \space n \in \Bbb{N}$$ The choice to use $(6n-5)$ was because I wanted to get rid of the $+5$ in the denominator, but I also needed to stick a coefficient on $n$ that was large enough where I could safely subtract $5$ at the same time and maintain the "less than or equal to" portion of the comparison test.
A: This is an overkill. All you need to show divergence is (for $a_k = \frac{5}{7k+28}$) is to consider the ratio
$$
\frac{a_k}{\frac{1}{k}} = \frac{5k}{7k +28} = \frac{5}{7}\cdot \frac{1}{1+\frac{28}{7k}} = \frac{5}{7}(1+o(1)) = O(1)
$$
Since $a_k = O \big(\frac{1}{k} \big)$ and $\sum_k \frac{1}{k}$ diverges, $\sum_k a_k$ diverges too.
