Absolute convergence: sum [n=1, infty] [(-1)^(n-1) 2^n/n^4]
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skeeter
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izzatso?
ratio test ...
\(\displaystyle \L \lim_{n\rightarrow \infty} \left| \frac{2^{n+1}}{(n+1)^4} \cdot \frac{n^4}{2^n} \right|\)
\(\displaystyle \L \lim_{n\rightarrow \infty} 2 \cdot \left( \frac{n}{n+1}\right)^4 = 2\)
since the limit > 1, the series diverges.
Since you know it converges, why not find out what to.
\(\displaystyle \L\\\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{4}}\)
\(\displaystyle \L\\= \;\ -1+\frac{1}{16}-\frac{1}{81}+\frac{1}{256}-\frac{1}{625}+\frac{1}{1296}-...............\)
Notice, the odds are negative and the evens are positive.
ODDS: \(\displaystyle \L\\\displaystyle{-}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{4}}=\frac{{-\pi}^{4}}{96}\)
EVENS: \(\displaystyle \L\\\displaystyle\sum_{k=1}^{\infty}\frac{1}{(2k)^{4}}=\frac{{\pi}^{4}}{1440}\)
\(\displaystyle \L\\\frac{{\pi}^{4}}{1440}-\frac{{\pi}^{4}}{96}=\frac{-7{\pi}^{4}}{720}\approx{-0.947032829497...}\)
\(\displaystyle \L\\\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{4}}\)
\(\displaystyle \L\\= \;\ -1+\frac{1}{16}-\frac{1}{81}+\frac{1}{256}-\frac{1}{625}+\frac{1}{1296}-...............\)
Notice, the odds are negative and the evens are positive.
ODDS: \(\displaystyle \L\\\displaystyle{-}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{4}}=\frac{{-\pi}^{4}}{96}\)
EVENS: \(\displaystyle \L\\\displaystyle\sum_{k=1}^{\infty}\frac{1}{(2k)^{4}}=\frac{{\pi}^{4}}{1440}\)
\(\displaystyle \L\\\frac{{\pi}^{4}}{1440}-\frac{{\pi}^{4}}{96}=\frac{-7{\pi}^{4}}{720}\approx{-0.947032829497...}\)