Infinite series
- Thread starter flakine
- Start date
Try the integral test.
\(\displaystyle \L\\\int_{1}^{\infty}\frac{4}{n(n+2)}dx=2ln(3)\)
The series converges by the integral test.
\(\displaystyle \frac{4}{1(1+2)}+\frac{4}{2(2+2)}+\frac{4}{3(3+2)}+....+\frac{4}{n(n+2)}\)
\(\displaystyle \L\\\frac{4}{3}+\frac{4}{8}+\frac{4}{15}+....+\frac{4}{n(n+2)}\)
\(\displaystyle \L\\4\left(\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+....+\frac{1}{n(n+2)}\right)\)
Can you find what it converges to?.
\(\displaystyle \L\\\int_{1}^{\infty}\frac{4}{n(n+2)}dx=2ln(3)\)
The series converges by the integral test.
\(\displaystyle \frac{4}{1(1+2)}+\frac{4}{2(2+2)}+\frac{4}{3(3+2)}+....+\frac{4}{n(n+2)}\)
\(\displaystyle \L\\\frac{4}{3}+\frac{4}{8}+\frac{4}{15}+....+\frac{4}{n(n+2)}\)
\(\displaystyle \L\\4\left(\frac{1}{3}+\frac{1}{8}+\frac{1}{15}+....+\frac{1}{n(n+2)}\right)\)
Can you find what it converges to?.
skeeter
Elite Member
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- Dec 15, 2005
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using the method of partial fractions ...
4/[n(n+2)] = 2/n - 2/(n+2) = 2[1/n - 1/(n+2)]
list out a few terms of the series 2[1/n - 1/(n+2) ...
2[(1 - 1/3) + (1/2 - 1/4) + (1/3 - 1/5) + (1/4 - 1/6) + ...
notice the 0 sums ... -1/3 +1/3, -1/4 +1/4, ...
what's left? ... what is the sum?
4/[n(n+2)] = 2/n - 2/(n+2) = 2[1/n - 1/(n+2)]
list out a few terms of the series 2[1/n - 1/(n+2) ...
2[(1 - 1/3) + (1/2 - 1/4) + (1/3 - 1/5) + (1/4 - 1/6) + ...
notice the 0 sums ... -1/3 +1/3, -1/4 +1/4, ...
what's left? ... what is the sum?