# Thread: Math Help Right Away!!!

1. ## Math Help Right Away!!!

2. (n!)/(n-2!) n= 2,3,4... Converges or Diverges? If converges where does it converge?

3. Summation of n=0 goes to infinity of the equation 2(-1/2)^n Converges where?

2. Originally Posted by Nimmy
2. (n!)/(n-2!) n= 2,3,4... Converges or Diverges? If converges where does it converge?

3. Summation of n=0 goes to infinity of the equation 2(-1/2)^n Converges where?
Hello,

to 2.) I hope you meant $\frac{n!}{(n-2)!}$. Then you can transform this term to:
$\frac{2\cdot3\cdot4\cdot...\cdot (n-2) \cdot (n-1)\cdot n}{2 \cdot3 \cdot 4 \cdot...\cdot (n-2)}$
and thats (n-1)*n which doesn't converge for n to infinity.

to 3. Split the summation into two sums: In 1st sum n is even and the numbers are positive, in the 2nd sum n is odd and all numbers are negative. Calculate both sums separately.

This sum must converge according to the convergence criterium of Leibniz.

Greetings

EB

3. The series,
$\sum^{\infty}_{k=0}2(-1/2)^k$.
Is the same as $2\sum^{\infty}_{k=0}(-1/2)^k$. Further, it is a geometric series with constant ratio $-1/2$. Thus, not only it converges we also can easily determine to what it converges to, which is $3/2$.

4. Originally Posted by ThePerfectHacker
The series,
$\sum^{\infty}_{k=0}2(-1/2)^k$.
Is the same as $2\sum^{\infty}_{k=0}(-1/2)^k$. Further, it is a geometric series with constant ratio $-1/2$. Thus, not only it converges we also can easily determine to what it converges to, which is $3/2$.
Ok Let me rephrase the question just to make sure the ppl in the fourm understand

Find the sum of the geometric series:

$\sum^{\infty}_{k=0}2(-1/2)^k$= 2-1+1/2-1/4...

can u explain how you got it to converge at 3/2?

5. Originally Posted by Nimmy
Ok Let me rephrase the question just to make sure the ppl in the fourm understand

Find the sum of the geometric series:

$\sum^{\infty}_{k=0}2(-1/2)^k$= 2-1+1/2-1/4...

can u explain how you got it to converge at 3/2?
Given the series,
$\sum^{\infty}_{k=0}r^k$
It converges if and only if, $|r|<1$.
Now if $|r|<1$ then it converges to,
$\frac{1}{1-r}$.
That is the property of the infinite geometric series.