# How do you use the Integral test on the infinite series sum_(n=1)^oo1/(2n+1)^3 ?

Sep 19, 2014

Since the corresponding integral
${\int}_{1}^{\infty} \frac{1}{2 x + 1} ^ 3 \mathrm{dx}$ converges to $\frac{1}{36}$,
the series
${\sum}_{n = 1}^{\infty} \frac{1}{2 n + 1} ^ 3$ also converges by Integral Test.

Let us evaluate the integral.

${\int}_{1}^{\infty} \frac{1}{2 x + 1} ^ 3 \mathrm{dx} = {\int}_{1}^{\infty} {\left(2 x + 1\right)}^{- 3} \mathrm{dx}$

Let $u = 2 x + 1$. Rightarrow {du}/{dx}=2 Rightarrow dx={du}/2
$x : 1 \to \infty R i g h t a r r o w u : 3 \to \infty$

$= {\int}_{3}^{\infty} {u}^{- 3} \frac{\mathrm{du}}{2}$

$= \frac{1}{2} {\lim}_{t \to \infty} {\int}_{3}^{t} {u}^{- 3} \mathrm{du}$

$= \frac{1}{2} {\lim}_{t \to \infty} {\left[{u}^{- 2} / \left\{- 2\right\}\right]}_{3}^{t}$

$= - \frac{1}{4} {\lim}_{t \to \infty} \left(\frac{1}{t} ^ 2 - \frac{1}{3} ^ 2\right)$

$= - \frac{1}{4} \left(0 - \frac{1}{9}\right)$

$= \frac{1}{36}$