Is it possible to know whether the following series converges without its formula?

Question

Given this series: $$\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{10} + \dfrac{1}{14} + \dfrac{1}{18} + \ldots$$ Is it possible to know if the series converges without its formula? I tried using the comparison test with both $1/n$ and $1/n^2$ but these don't yield results. The formula for this series is $1/[2(2n+1)]$, which I can use in other convergence tests, but if I didn't know that or couldn't find it, is it still possible to determine convergence without the formula?

Thanks

Answer 1

The formula is pretty easy as the denominators are in arithmetic sequence with common difference equal to four, so:

$$a_1=2\;,\;\;d=4\implies a_n=a_1+(n-1)d=2+4(n-1)=4n-2$$

and thus your series is simply

$$\sum_{n=1}^\infty\frac1{4n-2}$$

Now it is very simple to show the series is divergent.

Answer 2

$$\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{10} + \dfrac{1}{14} + \dfrac{1}{18} + \ldots > \\ \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{12} + \dfrac{1}{16} + \dfrac{1}{20} + \ldots = \\ \dfrac{1}{4}\left(\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \ldots \right)$$

Answer 3

Let $n\ge 1,$ integer.

$a_n = \dfrac{1}{2+(n-1)4} \gt $

$\dfrac{1}{4+4(n-1)} = \dfrac{1}{4n}$,

in essence the harmonic series, divergent.