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Series
Find the largest value of p for which the series diverges. Image: http://postimg.org/image/fjwytwxu7/
最佳解答:
case p > 0: f(n)= ln(n)/n^(p+1), f'(n) < 0, then f(n) is decreasing. by the integral test Σ(k=2,inf) ln(n)/n^(p+1) ~ integrate(x=2,inf) ln(x)/x^(p+1) dx = -(p lnx +1)/(p^2 x^p) for x=2, inf = (p ln2 +1)/(p^2 * 2^p) conv. case p < =0: Σ(k=2,inf) ln(n)/n^(p+1) > Σ(k=2,inf) ln(n)/n ~ integrate ln(x)/x dx = inf. div. Ans. p=0
其他解答:
Series
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發問:Find the largest value of p for which the series diverges. Image: http://postimg.org/image/fjwytwxu7/
最佳解答:
case p > 0: f(n)= ln(n)/n^(p+1), f'(n) < 0, then f(n) is decreasing. by the integral test Σ(k=2,inf) ln(n)/n^(p+1) ~ integrate(x=2,inf) ln(x)/x^(p+1) dx = -(p lnx +1)/(p^2 x^p) for x=2, inf = (p ln2 +1)/(p^2 * 2^p) conv. case p < =0: Σ(k=2,inf) ln(n)/n^(p+1) > Σ(k=2,inf) ln(n)/n ~ integrate ln(x)/x dx = inf. div. Ans. p=0
其他解答:
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