# Absolute convergence of a Serie

• Nov 11th 2008, 10:49 AM
amiv4
Absolute convergence of a Serie
Determine whether the series is absolutely convergent.

the series from 1 to infinity of

10^n/((n+1)4^(2n+1))
• Nov 11th 2008, 11:24 AM
Mathstud28
Quote:

Originally Posted by amiv4
Determine whether the series is absolutely convergent.

the series from 1 to infinity of

10^n/((n+1)4^(2n+1))

There is no alternating term here, so obviously if this fails to converge it does not converge absolutely.

Note $\displaystyle \sum_{n=1}^{\infty}\frac{10^n}{(n+1)4^{2n+1}}=\sum _{n=1}^{\infty}\frac{10^n}{4(n+1)16^n}\leq\sum_{n= 1}^{\infty}\left(\frac{10}{16}\right)^n=\frac{8}{3 }$
• Nov 11th 2008, 11:53 AM
amiv4
I'm not seeing what you did when u changed the original problem to that equation with the 16^n on the bottom. could u explain how u got that
• Nov 11th 2008, 11:54 AM
Mathstud28
Quote:

Originally Posted by amiv4
I'm not seeing what you did when u changed the original problem to that equation with the 16^n on the bottom. could u explain how u got that

$\displaystyle 4^{2n+1}=4\cdot{4^{2n}}=4\cdot\left(4^2\right)^n=4 \cdot{16^n}$