is series 1/(2^n -n) convergen

Set T(n) = 1/(2^n - n), compare T(n) and T(n+1). For n>0,if T(n+1) - T(n) >= 0, then diverge; if <= 0, then converge.
T(n+1) - T(n)= 1/[2^(n+1) - (n+1)] - 1/(2^n - n)= [2^n - n - 2^(n+1) + n + 1] / [2^(n+1) - (n+1)](2^n - n)= (1 - 2^n) / [2^(n+1) - (n+1)](2^n - n)= (-ve) / (+ve)(+ve)< 0Therefore, this series is converge.

2014-01-27 12:17:33 補充：
是的，感謝貓老師的提醒。
上面的理論是基於 T(n) > 0 for all positive integers n.
若果 not always positive, 那要證明 absolute of T(n)/T(n+1) < 1 就會 converge.

2014-01-27 14:02:26 補充：
The series T(n) = 1/n is converge, but your example is S(n).

2014-01-27 14:09:36 補充：
㢢，係咪我搞錯咗 series 同 sequence?
貓 Sir, 你答啦，我刪除。

how can it tends to 1/2?
n/2^n and (n+1)/2^n is not 0 as n tends to infinity

可能解釋方面需要作出些少調整。

例如： T(n) = -n

T(n+1) - T(n) = -1 < 0

but T(n) does not converge.

2014-01-27 12:04:04 補充：
How about using the ratio test?

http://en.wikipedia.org/wiki/Ratio_test

T(n) = 1/(2^n - n)

T(n+1)/T(n) = (2^n - n)/[2^(n+1) -n-1] = (1 - n/2^n)/[2 - (n+1)/2^n] → 1/2 as n→∞

Therefore, 1/(2^n - n) is a convergent series.

2014-01-27 12:33:17 補充：
Chun Yin (｡◕‿◕｡)

n/2^n → 0 as n→∞

Consider a real function f(x) = x/2^x.

lim(x→∞) x/2^x = lim(x→∞) 1/[2^x ln2] = 0 by L'Hopital rule.

Therefore, the limit of sequence n/2^n → 0 as n→∞.

2014-01-27 12:35:04 補充：
阿年　ღ(｡◕‿◠｡)ღ

Consider T(n) = 1/n > 0 for all n.

T(n+1) - T(n) = 1/(n+1) - 1/n = -1/[n(n+1)] < 0

However, 1/1 + 1/2 + 1/3 + 1/4 + ... does not converge.

2014-01-27 15:15:08 補充：
不如等多Ｄ意見先～

我唔係太熟 calculus...