convergent or divergent: sum [3^n (n!)^3] / [(3n)!]
- Thread starter synx
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skeeter
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let's try the ratio test again ...
all terms are > 0, so I'll dispense with the absolute value notation.
\(\displaystyle \L \frac{3^{n+1} [(n+1)!]^3}{[3(n+1)]!} \cdot \frac{(3n)!}{3^n (n!)^3} =\)
\(\displaystyle \L \frac{3(n+1)^3 (3n)!}{(3n+3)!} =\)
\(\displaystyle \L \frac{3(n+1)^3}{(3n+3)(3n+2)(3n+1)} =\)
\(\displaystyle \L \frac{3n^3 + 9n^2 + 9n + 3}{27n^3+54n^2+33n+6}\)
the limit of the last expession as n -> infinity would be 1/9 < 1, so the series converges.
all terms are > 0, so I'll dispense with the absolute value notation.
\(\displaystyle \L \frac{3^{n+1} [(n+1)!]^3}{[3(n+1)]!} \cdot \frac{(3n)!}{3^n (n!)^3} =\)
\(\displaystyle \L \frac{3(n+1)^3 (3n)!}{(3n+3)!} =\)
\(\displaystyle \L \frac{3(n+1)^3}{(3n+3)(3n+2)(3n+1)} =\)
\(\displaystyle \L \frac{3n^3 + 9n^2 + 9n + 3}{27n^3+54n^2+33n+6}\)
the limit of the last expession as n -> infinity would be 1/9 < 1, so the series converges.
Hello, synx!
Ratio Test: \(\displaystyle \L\:\frac{3^{n+1}\left[(n+1)!\right]^3}{(3n+3)!}\,\cdot\frac{(3n)!}{3^n(n!)^3} \;=\;\frac{3^{n+1}}{3^n}\,\cdot\,\frac{\left[(n+1)!\right]^3}{\left[n!\right]^3}\,\cdot\,\frac{(3n)!}{(3n+3)!}\)
. . \(\displaystyle \L=\;\frac{3}{1}\,\cdot\,\left[\frac{(n+1)!}{n!}\right]^3\cdot\,\frac{(3n!)}{(3n+3)(3n+2)(3n+1)(3n)!} \;=\;\frac{3}{1}\,\cdot\,\frac{(n+1)^3}{1}\,\cdot\,\frac{1}{(3n\,+\,3)(3n\,+\,2)(3n\,+\,1)}\)
. . \(\displaystyle \L=\;\frac{\not{3}(\sout{n\,+\,1})(n\,+\,1)(n\,+\,1)}{\not{3}(\sout{n\,+\,1})(3n\,+\,2)(3n\,+\,1)} \;=\;\frac{(n\,+\,1)(n\,+\,1)}{3n\,+\,2)(3n\,+\,1)}\)
Divide top and bottom by \(\displaystyle n^2:\) \(\displaystyle \L\:\frac{\left(\frac{n\,+\,1}{n}\right)\left(\frac{n\,+\,1}{n}\right)}{\left(\frac{3n\,+\,2}{n}\right)\left(\frac{3n\,+\,1}{n}\right)} \;=\;\frac{\left(1\,+\,\frac{1}{n}\right)\left(1\,+\,\frac{1}{n}\right)}{\left(3\,+\,\frac{2}{n}\right)\left(3\,+\,\frac{1}{n}\right)}\)
Take limit: \(\displaystyle \L\:\lim_{n\to\infty}\frac{\left(1\,+\,\frac{1}{n}\right)\left(1\,+\,\frac{1}{n}\right)}{\left(3\,+\,\frac{2}{n}\right)\left(3\,+\,\frac{1}{n}\right)} \;=\;\frac{(1)(1)}{(3)(3)} \;=\;\frac{1}{9}\)
Ratio Test: \(\displaystyle \L\:\frac{3^{n+1}\left[(n+1)!\right]^3}{(3n+3)!}\,\cdot\frac{(3n)!}{3^n(n!)^3} \;=\;\frac{3^{n+1}}{3^n}\,\cdot\,\frac{\left[(n+1)!\right]^3}{\left[n!\right]^3}\,\cdot\,\frac{(3n)!}{(3n+3)!}\)
. . \(\displaystyle \L=\;\frac{3}{1}\,\cdot\,\left[\frac{(n+1)!}{n!}\right]^3\cdot\,\frac{(3n!)}{(3n+3)(3n+2)(3n+1)(3n)!} \;=\;\frac{3}{1}\,\cdot\,\frac{(n+1)^3}{1}\,\cdot\,\frac{1}{(3n\,+\,3)(3n\,+\,2)(3n\,+\,1)}\)
. . \(\displaystyle \L=\;\frac{\not{3}(\sout{n\,+\,1})(n\,+\,1)(n\,+\,1)}{\not{3}(\sout{n\,+\,1})(3n\,+\,2)(3n\,+\,1)} \;=\;\frac{(n\,+\,1)(n\,+\,1)}{3n\,+\,2)(3n\,+\,1)}\)
Divide top and bottom by \(\displaystyle n^2:\) \(\displaystyle \L\:\frac{\left(\frac{n\,+\,1}{n}\right)\left(\frac{n\,+\,1}{n}\right)}{\left(\frac{3n\,+\,2}{n}\right)\left(\frac{3n\,+\,1}{n}\right)} \;=\;\frac{\left(1\,+\,\frac{1}{n}\right)\left(1\,+\,\frac{1}{n}\right)}{\left(3\,+\,\frac{2}{n}\right)\left(3\,+\,\frac{1}{n}\right)}\)
Take limit: \(\displaystyle \L\:\lim_{n\to\infty}\frac{\left(1\,+\,\frac{1}{n}\right)\left(1\,+\,\frac{1}{n}\right)}{\left(3\,+\,\frac{2}{n}\right)\left(3\,+\,\frac{1}{n}\right)} \;=\;\frac{(1)(1)}{(3)(3)} \;=\;\frac{1}{9}\)