Alternating Series Problem
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BigGlenntheHeavy
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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.
\(\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k+1}k!}{(2k-1)!}\)
Hint: Use the ratio test, and you'll find that the sum convergerces absolutely.
pka
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\(\displaystyle \left[ {\frac{{\left( {k + 1} \right)!}}{{\left( {2\left( {k + 1} \right) - 1} \right)!}}} \right]\left[ {\frac{{k!}}{{\left( {2k - 1} \right)!}}} \right]^{ - 1} = \frac{{\left[ {\left( {k + 1} \right)!} \right]\left( {2k - 1} \right)!}}{{\left( {2k + 1} \right)!\left( k \right)!}} = \frac{{k + 1}}{{\left( {2k + 1} \right)\left( {2k} \right)}}\)
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
\(\displaystyle \boxed{-\frac{k+1}{2k(2k+1)}}\)
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
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\(\displaystyle [\frac{(k+1)!}{(2k+1)!}][\frac{(2k-1)!}{k!}] \ = [\ {\frac{(k+1)k!}{(2k+1)(2k)(2k-1)!}][\frac{(2k-1)!}{k!}]\)
\(\displaystyle \ = \frac{(k+1)}{(2k+1)(2k)} \ Hence, \ as \ k \ approaches \ infinity, \ the \ limit \ approaches \ 0 \ which \ is \ less \ than \ 1.\)
\(\displaystyle Ergo, \ the \ summation \ converges \ absolutely.\)
Note: the absolute value of (-1)^(k+1) is one ,so we can disregard.
\(\displaystyle \ = \frac{(k+1)}{(2k+1)(2k)} \ Hence, \ as \ k \ approaches \ infinity, \ the \ limit \ approaches \ 0 \ which \ is \ less \ than \ 1.\)
\(\displaystyle Ergo, \ the \ summation \ converges \ absolutely.\)
Note: the absolute value of (-1)^(k+1) is one ,so we can disregard.