How do you find the 5-th partial sum of the infinite series $sum_(n=1)^oo1/(n(n+2)$ ?

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How do you find the 5-th partial sum of the infinite series $sum_(n=1)^oo1/(n(n+2)$ ?

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Answer 1 · 2014-08-30T01:40:03

This is simply the sum of the first 5 terms, so we have
$S_5=1/{1(1+2)}+1/{2(2+2)}+1/{3(3+2)}+1/{4(4+2)}+1/{5(5+2)} =1/3+1/8+1/15+1/24+1/35approx0.6$ .

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How do you find the 5-th partial sum of the infinite series $sum_(n=1)^oo1/(n(n+2)$ ?

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How do you find the 5-th partial sum of the infinite series sum_(n=1)^oo1/(n(n+2)sum_(n=1)^oo1/(n(n+2) ?

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How do you find the 5-th partial sum of the infinite series $sum_(n=1)^oo1/(n(n+2)$ ?