搜搜做题作业网Does the series converge?If it does,try to find the sum
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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
∞∑(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
再问： =1/2*(1-(1/2)^n)÷ (1-1/2)-n/2^(n+1) 这个式子缺括号不？ 怎么感觉看的迷糊 还有 3楼得答案跟你 接近 不过他是S=2-1/2^(n-1)-n/2^(n+1) 这样= = 很纠结
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
再问： =1/2*(1-(1/2)^n)÷ (1-1/2)-n/2^(n+1) 这个式子缺括号不？ 怎么感觉看的迷糊 还有 3楼得答案跟你 接近 不过他是S=2-1/2^(n-1)-n/2^(n+1) 这样= = 很纠结
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