sin(n^2 an n)
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很简单(sinn)/n^2≤1/n^2因为|sinn|≤1∑1/n^2绝对收敛,所以原级数也绝对收敛
分子=sin(n+1)A+2sin(n)A+sin(n-1)A=[sin(n+1)A+sinnA]+[sinnA+sin(n-1)A]=2sin(2n+1)A/2*cosA/2+2sin(2n-1)A
lim(n→∞)[(sin(2/n)+cos(3/n))^(-n)]=lim(n→∞)[(sin(2/n)+1)^(-n)]=e^[lim(n→∞)(-n)ln(sin(2/n)+1)](等价无穷小替
n→∞,1/n→0+,所以可以令x=1/n→0+后,两极限是等价的(由海因定理保证)lim(1/n-sin(1/n))/(1/n^2)=lim(x-sinx)/(x^2),和lim(1/n-sin(1
用复数w=cos(2π/n)+isin(2π/n)w'=cos(2π/n)-isin(2π/n)z^n=1(z-1)(z^(n-1)+z^(n-2)+……+z+1)=0z^(n-1)+z^(n-2)+
不确定你是从哪里得来的词,今天上班问了下德国同事,大家都没有见过这个词,Ann是人名,如果你是听写的话,有可能是einGlück.
n->∞,1/(n+2)->0所以sin(1/(n+2))等价于1/(n+2)[1/(n+2)]/[3/(2n+5)]=(2n+5)/(3n-6)->2/3所以是同阶但非等阶的无穷小注意:两个式子比值
该级数实为1,0,-1/3,0,1/5,0,-1/7,0,……,1/4t,0,-1/(4t+2),0,……我们将1/4t,0,-1/(4t+2),0的和组成一项有an=1/4n-1/(4n+2)=1/
(n^2+2)^0.5=n+2/((n^2+2)^0.5+n),为方便,记2/((n^2+2)^0.5+n)=t.sin(π(n^2+2)^0.5)=sin(π(n+t))=(-1)^(n-1)*si
应用数学归纳法.1.当n=1时,左边=sin(pi/3),右边=sin(pi/3).则命题成立2.假设当n=k时,命题成立.即sin(pi/3)+sin(2*pi/3)+...+sin(k*pi/3)
a2-a1=2,a3-a2=4,…an+1-an=2n,这n个式子相加,就有an+1=16+n(n+1),即an=n(n-1)+16=n2-n+16,∴ann=n+16n−1,用均值不等式,知道它在n
积化和差,sin(a+b)=sinacosb+cosasinb,1sin(a-b)=sinacosb-cosasinb,21-2,sin(a+b)-sin(a-b)=2sinacosb;令a=mx,b
这是收敛的lim(n->inf)π/n=0lim(sin(pai/n)^2)=sin(lim(n->inf)π/n)^2=0所以从结果看来,是收敛的.
n→∞,2nπ/(3n+1)→2π/3∴0<sin(2nπ/(3n+1))→√3/2<1∴[sin(2nπ/(3n+1)]^n→0
sinx-2/Pi*x这个函数,在0和Pi/2都等于0,并且在这个区间上是凹函数,所以大于等于0.
lim【n→∞】(2n²-3n+1)/(n+1)×sin(1/n)=lim【n→∞】(2n²-3n+1)/(n+1)×(1/n)=lim【n→∞】(2n²-3n+1)/(
关于n的数列极限问题,可以转化为函数极限:n^2*ln[n*sin(1/n)]=【ln{[sin(1/n)]/(1/n)}】/[(1/n)^2]当n→+∞时,1/n→0,所以用x代替式中的1/n得到:
由题意可得an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=2(n-1)+2(n-2)+…+2+33=[2(n−1)+2](n−1)2+33=n2-n+33,故ann=n2−