高数:无穷级数判别敛散性1+(1+2)/(1+2^2)+(1+3)/(1+3^2)+...+(1+n)/(1+n^2)+...;为什么(1+n)/(1+n^2)> (1+n)/(1+n^2+2n)?1+n^2
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![高数:无穷级数判别敛散性1+(1+2)/(1+2^2)+(1+3)/(1+3^2)+...+(1+n)/(1+n^2)+...;为什么(1+n)/(1+n^2)> (1+n)/(1+n^2+2n)?1+n^2](/uploads/image/z/1825693-61-3.jpg?t=%E9%AB%98%E6%95%B0%EF%BC%9A%E6%97%A0%E7%A9%B7%E7%BA%A7%E6%95%B0%E5%88%A4%E5%88%AB%E6%95%9B%E6%95%A3%E6%80%A71%2B%EF%BC%881%2B2%EF%BC%89%2F%281%2B2%5E2%29%2B%281%2B3%29%2F%281%2B3%5E2%29%2B...%2B%281%2Bn%29%2F%281%2Bn%5E2%29%2B...%3B%E4%B8%BA%E4%BB%80%E4%B9%88%281%2Bn%29%2F%281%2Bn%5E2%29%3E+%281%2Bn%29%2F%281%2Bn%5E2%2B2n%EF%BC%89%EF%BC%9F1%2Bn%5E2)
高数:无穷级数判别敛散性1+(1+2)/(1+2^2)+(1+3)/(1+3^2)+...+(1+n)/(1+n^2)+...;为什么(1+n)/(1+n^2)> (1+n)/(1+n^2+2n)?1+n^2
高数:无穷级数判别敛散性
1+(1+2)/(1+2^2)+(1+3)/(1+3^2)+...+(1+n)/(1+n^2)+...;
为什么(1+n)/(1+n^2)> (1+n)/(1+n^2+2n)?
1+n^2
高数:无穷级数判别敛散性1+(1+2)/(1+2^2)+(1+3)/(1+3^2)+...+(1+n)/(1+n^2)+...;为什么(1+n)/(1+n^2)> (1+n)/(1+n^2+2n)?1+n^2
∵Un=(1+n)/(1+n^2)>1/n
∵∑1/n是发散的
∴∑(1+n)/(1+n^2)是发散的.
这个级数是发散的!
首先:1+1/2+1/3+.....+1/n 这个级数是发散的,教材上一般都列举了这个例子的!
在看你的题目中,(1+n)/(1+n^2)> (1+n)/(1+n^2+2n) = 1/(n+1)
所以:1+(1+2)/(1+2^2)+(1+3)/(1+3^2)+...+(1+n)/(1+n^2)+...; > 1+1/2+1/3+.....+1/n +...
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这个级数是发散的!
首先:1+1/2+1/3+.....+1/n 这个级数是发散的,教材上一般都列举了这个例子的!
在看你的题目中,(1+n)/(1+n^2)> (1+n)/(1+n^2+2n) = 1/(n+1)
所以:1+(1+2)/(1+2^2)+(1+3)/(1+3^2)+...+(1+n)/(1+n^2)+...; > 1+1/2+1/3+.....+1/n +1/(1+n)
故:上面级数是发散的!
收起
n趋于无穷时(1+n)/(1+n^2)与1/n同阶,1+1/2+……+(1/n)+……与原级数同敛散。由于调和级数1+1/2+……+(1/n)+……发散,故原级数发散
正项数列{An}、{Bn}如果满足n趋于无穷时lim(An/Bn)=l>0,则{An}、{Bn}相对应的级数同敛散。
利用极限审敛法,n*(1+n)/(1+n^2)=(n^2+n)/(1+n^2)上式右端趋于正无穷时极限为1,根据极限审敛法,题设级数发散