求证求和级数(-1)^[n^0.5]/n收敛,其中[n^0.5]的意思是n开根号然后取证取整 (-1)^[n^0.5]再乘以1/n
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求证求和级数(-1)^[n^0.5]/n收敛,其中[n^0.5]的意思是n开根号然后取证取整 (-1)^[n^0.5]再乘以1/n
求证求和级数(-1)^[n^0.5]/n收敛,其中[n^0.5]的意思是n开根号然后取证取整 (-1)^[n^0.5]再乘以1/n
求证求和级数(-1)^[n^0.5]/n收敛,其中[n^0.5]的意思是n开根号然后取证取整 (-1)^[n^0.5]再乘以1/n
设int(√n)=i(int为取整运算),i=1、2、3、...为正整数,满足int(√n)=i的n值有2i+1个：i^2、i^2+1、i^2+2、...、i^2+2i；可以将这2i+1项合并为1项,显然这2i+1项具有相同的符号；令a(i)=(-1)^i[1/i^2+1/(i^2+1)+1/(i^2+2)+...+1/(i^2+2i)]；原级数可以用a(i)表示；显然,a(i)为交错级数；|a(i)|=1/i^2+1/(i^2+1)+...+1/(i^2+2i)＜(2i+1)/i^2,所以当i→+∞时,|a(i)|→0,即lim(i→+∞)|a(i)|=0①；另外|a(i)|=1/[i(i+1)]+∑(k=0→i-1)[1/(i^2+k)+1/(i^2+2i-k)]=1/[i(i+1)]+∑(k=0→i-1)2i(i+1)/(i^4+2i^3+2ki-k^2)②；由②得|a(i)|＞1/[i(i+1)]+∑(k=0→i-1)2(i+1)/(i^3+2i^2+2i-2),即|a(i)|＞1/[i(i+1)]+2i(i+1)/(i^3+2i^2+2i-2)③；由②得|a(i)|＜1/[i(i+1)]+2i^2(i+1)/[i^4+2i^3-(i-1)^2]④；由④得|a(i+1)|＜1/[(i+1)(i+2)]+2(i+1)^2(i+2)/[(i+1)^4+2(i+1)^3-i^2]⑤；由③、⑤得|a(i)|＞1/[i(i+1)]+2i(i+1)/[i^3+2i^2+2i-2]＞1/[(i+1)(i+2)]+2(i+1)^2(i+2)/[(i+1)^4+2(i+1)^3-i^2]＞|a(i+1)|,即|a(i)|＞|a(i+1)|⑥；由①、⑥及莱布尼兹准则可知,级数a(i)收敛,所以原级数收敛.