# converge or diverge

• Aug 15th 2009, 03:01 PM
johntuan
converge or diverge
Can you guys check my work to see if I did it right

the function is [(n^2)(2^n-1)]/(-5^n)

so what i did was use the nth root test and got (n^(2/n))/10..(is the lim of n^(2/n) equal to 1??..i just used my calculator for this..)

so the limit of that would be 1/10 <1 therefore converge

am i right?
• Aug 15th 2009, 03:21 PM
Plato
Quote:

Originally Posted by johntuan
the function is [(n^2)(2^n-1)]/(-5^n)

There are some real notational problems to this question.

$\displaystyle -5^n$ is a negative number. $\displaystyle \sum {\frac{1}{{ - 5^n }} = - \sum {\frac{1}{{5^n }}} }$.

Do you mean $\displaystyle (-5)^n$?

Is it $\displaystyle 2^n-1$ or $\displaystyle 2^{n-1}?$
• Aug 15th 2009, 03:25 PM
johntuan
Quote:

Originally Posted by Plato
There are some real notational problems to this question.

$\displaystyle -5^n$ is a negative number. $\displaystyle \sum {\frac{1}{{ - 5^n }} = - \sum {\frac{1}{{5^n }}} }$.

Do you mean $\displaystyle (-5)^n$?

Is it $\displaystyle 2^n-1$ or $\displaystyle 2^{n-1}$

for the (-5), I thought it didnt matter because when using the nth-root test, its the absolute value that's using.

so yeah i meant $\displaystyle (-5)^n$

and $\displaystyle 2^{n-1}?$ is the correct one.

sorry for the confusion
• Aug 15th 2009, 03:27 PM
Twig
Hi

I assume you mean $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}n^{2}(2^{n}-1)}{5^{n}}$

$\displaystyle |\frac{(-1)^{n}n^{2}(2^{n}-1)}{5^{n}}|\leq \frac{n^{2}(2^{n})}{5^{n}}$

So this series is absolute convergent.
• Aug 15th 2009, 03:44 PM
johntuan
Quote:

Originally Posted by Twig
Hi

I assume you mean $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}n^{2}(2^{n}-1)}{5^{n}}$

$\displaystyle |\frac{(-1)^{n}n^{2}(2^{n}-1)}{5^{n}}|\leq \frac{n^{2}(2^{n})}{5^{n}}$

So this series is absolute convergent.

no theres no (-1)^n in the numerator and (-5^n) on the denominator