T*******x 发帖数: 8565 | 1 给定任意正整数 n>1,
证明级数收敛:
sum (k from 0 to infinity) sum (m from 0 to n-1) of
exp(2*pi*i*m/n) * 1/(k*n+m+1) |
T*******x 发帖数: 8565 | 2 相关的一个问题:
integration x from 1 to infinity of
(1/x) * exp(ix/L)
收敛for sufficiently large L.
【在 T*******x 的大作中提到】![](/moin_static193/solenoid/img/up.png) : 给定任意正整数 n>1, : 证明级数收敛: : sum (k from 0 to infinity) sum (m from 0 to n-1) of : exp(2*pi*i*m/n) * 1/(k*n+m+1)
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w****n 发帖数: 113 | 3 This is too trivial. When k is large, 1/(kn+m+1)=1/(kn)+O((m+1)/(kn)^2).
Thus the innermost sum is 1/(kn)sum_{m}e(m/n)+sum_{m}O((m+1)/(kn)^2)=O(1/k^2
).
【在 T*******x 的大作中提到】![](/moin_static193/solenoid/img/up.png) : 给定任意正整数 n>1, : 证明级数收敛: : sum (k from 0 to infinity) sum (m from 0 to n-1) of : exp(2*pi*i*m/n) * 1/(k*n+m+1)
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T*******x 发帖数: 8565 | 4 嗯。正确。
^2
【在 w****n 的大作中提到】![](/moin_static193/solenoid/img/up.png) : This is too trivial. When k is large, 1/(kn+m+1)=1/(kn)+O((m+1)/(kn)^2). : Thus the innermost sum is 1/(kn)sum_{m}e(m/n)+sum_{m}O((m+1)/(kn)^2)=O(1/k^2 : ).
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