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

Question

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

Paisley Johnson · Accepted Answer

Since the corresponding integral
#int_1^infty 1/(2x+1)^3dx# converges to #1/36#,
the series
#sum_{n=1}^infty1/(2n+1)^3# also converges by Integral Test. Let us evaluate the integral. #int_1^infty 1/(2x+1)^3dx=int_1^infty(2x+1)^{-3}dx# Let #u=2x+1#. #Rightarrow {du}/{dx}=2 
Rightarrow dx={du}/2#
#x: 1 to infty Rightarrow u: 3 to infty# #=int_3^infty u^{-3}{du}/2# #=1/2lim_{t to infty}int_3^t u^{-3}du# #=1/2lim_{t to infty}[u^{-2}/{-2}]_3^t# #=-1/4lim_{t to infty}(1/t^2-1/3^2)# #=-1/4(0-1/9)# #=1/36#

Christopher Allers · Answer

undefined To use the Integral Test on the infinite series $ \sum_{n=1}^{\infty} \frac{1}{(2n+1)^3} $, you need to:

1. Define the function $ f(x) = \frac{1}{(2x+1)^3} $.
2. Check if $ f(x) $ is continuous, positive, and decreasing for $ x \geq 1 $.
3. Integrate $ f(x) $ from 1 to $ \infty $.
4. If the integral converges, then the series converges. If the integral diverges, then the series also diverges.

Now, integrating $ f(x) $:

$$ \int_1^{\infty} \frac{1}{(2x+1)^3} \, dx $$

$$ \text{Let } u = 2x+1, \, du = 2 \, dx $$

$$ \frac{1}{2} \int_{3}^{\infty} \frac{1}{u^3} \, du $$

$$ \frac{1}{2} \left[ -\frac{1}{2u^2} \right]_{3}^{\infty} $$

$$ \frac{1}{4} \left( 0 - \left( -\frac{1}{18} \right) \right) $$

$$ \frac{1}{72} $$

Since the integral converges, by the Integral Test, the series $ \sum_{n=1}^{\infty} \frac{1}{(2n+1)^3} $ also converges.