# Thread: converge or diverge

1. ## 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?

2. Originally Posted by johntuan
the function is [(n^2)(2^n-1)]/(-5^n)
There are some real notational problems to this question.

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

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

Is it $2^n-1$ or $2^{n-1}?$

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

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

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

Is it $2^n-1$ or $2^{n-1}$
sorry about the notation problem

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 $(-5)^n$

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

sorry for the confusion

4. Hi

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

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

So this series is absolute convergent.

5. Originally Posted by Twig
Hi

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

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

So this series is absolute convergent.
no theres no (-1)^n in the numerator and (-5^n) on the denominator