級數收斂發散問題(高點數20)

1. div.
lim(n->∞) n^(1/n) = 1 , lim(n->∞) ln(n)=∞
so, there exists M>0 such that for all n>=M, n^(1/n) < ln(n)
then∑(n=M~∞) 1/n^(1+1/n) > ∑(n=M~∞) 1/(n lnn) > ∫[M~∞] 1/(x lnx) dx=∞
thus, ∑(n=1~∞) 1/n^(1+1/n) diverges.

2. conv. by the test for alternating series
(since, 1/ n^0.5 decreases to zero.)

3. ∑(n=2 ~∞) 1/{n*(ln n)^n}
∑(n=3~∞) 1/{ n*(ln n)^2} < ∑(n=3~∞) 1/{ n(ln n)^2}
<∫[2~∞] 1/[x (lnx)^2] dx = -1/ln(x) |[2~∞]
and 1/{ n (ln n)^n } decreases to zero,
thus, ∑(n=2~∞) 1/{ n*(ln n)^n } converges.

Note: when n=1, 1/{n*(ln n)^n }=∞,
so, ∑(n=1~∞) 1/[n (ln n)^n] is meaningless.

4. conv. by limit comparison test
lim(n->∞) arctan[ 1/(n^2+n+1)]/ (1 /n^2) = 1
so, ∑(n=1~∞) arctan[1/(n^2+n+1)] is similar to ∑(n=1~∞) 1/n^2
while ∑(n=1~∞) 1/n^2 is a conv. p-series (p=2)
thus, ∑(n=1~∞) arctan[1/(n^2+n+1)] conv.

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第四個，高中程度作法
arctan(1/(n^2+n+1))=arctan(1/n)-arctan(1/(n+1))