# How do you test the series Sigma 2/(4n-3) from n is [1,oo) for convergence?

Jan 18, 2018

We can apply d'Alembert's ratio test:

Suppose that;

$S = {\sum}_{r = 1}^{\infty} {a}_{n} \setminus \setminus$, and $\setminus \setminus L = {\lim}_{n \rightarrow \infty} | {a}_{n + 1} / {a}_{n} |$

Then

if L < 1 then the series converges absolutely;
if L > 1 then the series is divergent;
if L = 1 or the limit fails to exist the test is inconclusive.

So our series is;

$S = {\sum}_{n = 0}^{\infty} {a}_{n}$ where ${a}_{n} = \frac{2}{4 n - 3}$

So our test limit is:

$L = {\lim}_{n \rightarrow \infty} \left\mid \frac{\frac{2}{4 \left(n + 1\right) - 3}}{\frac{2}{4 n - 3}} \right\mid$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} \left\mid \frac{2}{4 n + 4 - 3} \cdot \frac{4 n - 3}{2} \right\mid$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} \left\mid \frac{4 n - 3}{4 n + 1} \right\mid$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} \left\mid \frac{\left(4 n + 1\right) - 4}{4 n + 1} \right\mid$
$\setminus \setminus \setminus = {\lim}_{n \rightarrow \infty} \left\mid 1 - \frac{4}{4 n + 1} \right\mid$
$\setminus \setminus \setminus = 1$

and we can conclude that in the case the ratio test is inconclusive, (in fact the series diverges)