作业帮Does the series converge?If it does,try to find the sum∞∑(n=1) n/2^n 题就是这样~∞在∑上面（n=1)在∑下面 右边是2的n次方分之n 如果解答过程有字的话最好是英文
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Does the series converge?If it does,try to find the sum∞∑(n=1) n/2^n 题就是这样~∞在∑上面（n=1)在∑下面 右边是2的n次方分之n 如果解答过程有字的话最好是英文
Does the series converge?If it does,try to find the sum
∞∑(n=1) n/2^n 题就是这样~∞在∑上面（n=1)在∑下面 右边是2的n次方分之n
如果解答过程有字的话最好是英文
Does the series converge?If it does,try to find the sum∞∑(n=1) n/2^n 题就是这样~∞在∑上面（n=1)在∑下面 右边是2的n次方分之n 如果解答过程有字的话最好是英文
设S=1/2+2/4+3/8+……+n/2^n
S/2=1/4+2/8+……+(n-1)/2^n+n/2^(n+1)
两式想减得：
S/2=1/2+1/4+1/8+……+1/2^n-n/2^(n+1)
=1/2*(1-(1/2)^n)÷ (1-1/2)-n/2^(n+1)
=1-1/2^n-n/2^(n+1)
∴ S=2-1/2^(n-1)-n/2^n
所以和的极限是2,翻译就算了吧,我不行啊
Un/U(n+1)=2-2/(n+2) 2>1 so the series converge .
U1=1/2
U(n+1)/Un = 1/2+1/(2n)
2U(n+1)=(1+1/n)Un=Un +(1/n)*(n/2^n) =Un+1/(2^n)
2S(n+1)-2U1 = Sn+(1/2+......+1/(2^n))=Sn+(1-1/(2^n))
Sn=2-(n+2)/(2^n)
晕死，书写太麻烦，你在核算一遍哈。
suppose S(k)=∑(n=1,k) (n/2^n).
S(k)/2=∑(n=1,k) (n/2^(n+1))=∑(n=2,k+1) [(n-1)/2^n]=-1/2+k/2^(k+1)+∑(n=1,k) (n/2^n)-∑(n=2,k+1) (1/2^n)=k/2^(k+1)+S(k)-∑(n=1,k+1) (1/2^n)=k/2^(k+1)-1+1/2^k.
Thus...
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suppose S(k)=∑(n=1,k) (n/2^n).
S(k)/2=∑(n=1,k) (n/2^(n+1))=∑(n=2,k+1) [(n-1)/2^n]=-1/2+k/2^(k+1)+∑(n=1,k) (n/2^n)-∑(n=2,k+1) (1/2^n)=k/2^(k+1)+S(k)-∑(n=1,k+1) (1/2^n)=k/2^(k+1)-1+1/2^k.
Thus, S(k)=2-1/2^(k-1)-k/2^(k+1).
Let k->∞, it is obvious that 1/2^(k-1)->0, and k/2^(k+1)->0. So we can conclude that ∑(n=1,∞) n/2^n converges, and the value is 2.
打数学式子太费劲了……
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