Alternating Series Problem

[hgaon001]

The question: Classify each series as absolutely convergent, conditionally convergent, or divergent.

The summation from k=1 to infinity of [(-1)^(k+1)] k!/(2k-1)!

 

[BigGlenntheHeavy]

Test for convergence and/or divergence.

$\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k+1}k!}{(2k-1)!}$

Hint: Use the ratio test, and you'll find that the sum convergerces absolutely.

 

[hgaon001]

Test for convergence and/or divergence.

$\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k+1}k!}{(2k-1)!}$

Hint: Use the ratio test, and you'll find that the sum convergerces absolutely.

The thing is that the answer i got was (k+1)(2k-1)/(2k+1) and the book gets (k+1)/(2k+1)(2k) and i cant figure out how they get the 2k without the factorial at the bottom.

 

[pka]

$\displaystyle \left[ {\frac{{\left( {k + 1} \right)!}}{{\left( {2\left( {k + 1} \right) - 1} \right)!}}} \right]\left[ {\frac{{k!}}{{\left( {2k - 1} \right)!}}} \right]^{ - 1} = \frac{{\left[ {\left( {k + 1} \right)!} \right]\left( {2k - 1} \right)!}}{{\left( {2k + 1} \right)!\left( k \right)!}} = \frac{{k + 1}}{{\left( {2k + 1} \right)\left( {2k} \right)}}$

 

[galactus]

It's a matter of manipulating the factorials algebraically.

By the ratio test, we get:

$\displaystyle \frac{(-1)^{k+2}(k+1)!}{(2(k+1)-1)!}\cdot\frac{(2k-1)!}{(-1)^{k+1}k!}$

Note that:

$\displaystyle (k+1)!=k!(k+1), \;\ (2(k+1)-1)!=(2k)!(2k+1), \;\ (2k-1)!=\frac{(2k)!}{2k}$

Put it altogether and note the cancellations:

$\displaystyle \frac{k!(k+1)}{(2k)!(2k+1)}\cdot\frac{\frac{(2k)!}{2k}}{k!}=\frac{k+1}{2k(2k+1)}$

Note that $\displaystyle \frac{(-1)^{k+2}}{(-1)^{k+1}}=-1$, we could have disregarded that. so that gives us

$\displaystyle \boxed{-\frac{k+1}{2k(2k+1)}}$

 

[BigGlenntheHeavy]

\(\displaystyle [\frac{(k+1)!}{(2k+1)!}][\frac{(2k-1)!}{k!}] \ = [\ {\frac{(k+1)k!}{(2k+1)(2k)(2k-1)!}][\frac{(2k-1)!}{k!}]\)

$\displaystyle \ = \frac{(k+1)}{(2k+1)(2k)} \ Hence, \ as \ k \ approaches \ infinity, \ the \ limit \ approaches \ 0 \ which \ is \ less \ than \ 1.$

$\displaystyle Ergo, \ the \ summation \ converges \ absolutely.$

Note: the absolute value of (-1)^(k+1) is one ,so we can disregard.