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?

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- Feb 21st 2006, 08:26 PMNimmyMath 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? - Feb 21st 2006, 08:54 PMearbothQuote:

Originally Posted by**Nimmy**

to 2.) I hope you meant $\displaystyle \frac{n!}{(n-2)!}$. Then you can transform this term to:

$\displaystyle \frac{2\cdot3\cdot4\cdot...\cdot (n-2) \cdot (n-1)\cdot n}{2 \cdot3 \cdot 4 \cdot...\cdot (n-2)}$

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 - Feb 22nd 2006, 02:23 PMThePerfectHacker
The series,

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

Is the same as $\displaystyle 2\sum^{\infty}_{k=0}(-1/2)^k$. Further, it is a geometric series with constant ratio $\displaystyle -1/2$. Thus, not only it converges we also can easily determine to what it converges to, which is $\displaystyle 3/2$. - Feb 23rd 2006, 03:16 PMNimmyQuote:

Originally Posted by**ThePerfectHacker**

Find the sum of the geometric series:

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

can u explain how you got it to converge at 3/2? - Feb 23rd 2006, 05:16 PMThePerfectHackerQuote:

Originally Posted by**Nimmy**

$\displaystyle \sum^{\infty}_{k=0}r^k$

It converges if and only if, $\displaystyle |r|<1$.

Now if $\displaystyle |r|<1$ then it converges to,

$\displaystyle \frac{1}{1-r}$.

That is the property of the infinite geometric series.