Is this series convergent or diver, conditionally convergent or absolutly convergent?
- Thread starter biogirl
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pka
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You series is such that \(\displaystyle a_n\le\dfrac{2\cdot 5^n}{n!}\).
Use the ratio test on that to get \(\displaystyle \dfrac{5}{n+1}\).
Now use simple comparison.
Hello, biogirl
We can use the Ratio Test . . . takes a bit of gymnastics.
. . \(\displaystyle \displaystyle R \;=\;\frac{5^{n+1}+(n+1)}{(n+1)!+3} \cdot\frac{n!+3}{5^n+n} \;=\;\frac{5^{n+1}+(n+1)}{5^n+n}\cdot\frac{n!+3}{(n+1)! + 3}\)
Divide numerator and denominator of the first fraction by \(\displaystyle 5^n\)
Divide numerator and denominator of the second fraction by \(\displaystyle (n+1)!\)
. . \(\displaystyle R \;=\; \dfrac{5 + \frac{n+1}{5^n}}{1 + \frac{n}{5^n}}\cdot \dfrac{\frac{1}{n+1} + \frac{3}{(n+1)!}}{1 + \frac{3}{(n+1)!}} \)
Then: .\(\displaystyle \displaystyle \lim_{n\to\infty} R \;=\;\lim_{n\to\infty}\frac{5 + \frac{n+1}{5^n}}{1 + \frac{n}{5^n}}\cdot \frac{\frac{1}{n+1} + \frac{3}{(n+1)!}}{1 + \frac{3}{(n+1)!}} \)
. . . . . . . . . . .\(\displaystyle =\;\dfrac{5+0}{1+0}\cdot \dfrac{0+0}{1+0} \;=\;5\cdot0\;=\;0\)
The sequence converges absolutely.
We can use the Ratio Test . . . takes a bit of gymnastics.
. . \(\displaystyle \displaystyle R \;=\;\frac{5^{n+1}+(n+1)}{(n+1)!+3} \cdot\frac{n!+3}{5^n+n} \;=\;\frac{5^{n+1}+(n+1)}{5^n+n}\cdot\frac{n!+3}{(n+1)! + 3}\)
Divide numerator and denominator of the first fraction by \(\displaystyle 5^n\)
Divide numerator and denominator of the second fraction by \(\displaystyle (n+1)!\)
. . \(\displaystyle R \;=\; \dfrac{5 + \frac{n+1}{5^n}}{1 + \frac{n}{5^n}}\cdot \dfrac{\frac{1}{n+1} + \frac{3}{(n+1)!}}{1 + \frac{3}{(n+1)!}} \)
Then: .\(\displaystyle \displaystyle \lim_{n\to\infty} R \;=\;\lim_{n\to\infty}\frac{5 + \frac{n+1}{5^n}}{1 + \frac{n}{5^n}}\cdot \frac{\frac{1}{n+1} + \frac{3}{(n+1)!}}{1 + \frac{3}{(n+1)!}} \)
. . . . . . . . . . .\(\displaystyle =\;\dfrac{5+0}{1+0}\cdot \dfrac{0+0}{1+0} \;=\;5\cdot0\;=\;0\)
The sequence converges absolutely.