# How do you test the series Sigma 1/(2^n-n) from n is [1,oo) for convergence?

Jan 8, 2017

The series converges by the ratio test.

#### Explanation:

We do a series ratio test

We calculate L=lim_(n->oo)∣a_(n+1)/a_n∣

Here,

∣a_(n+1)/a_n∣=∣(1/(2^(n+1)-(n+1)))/(1/(2^n-n))∣

=∣(2^n-n)/(2^(n+1)-(n+1))∣

Therefore,

L=lim_(n->oo)∣a_(n+1)/a_n∣=lim_(n->oo)∣(2^n-n)/(2^(n+1)-(n+1))∣

$= {\lim}_{n \to \infty} {2}^{n} / \left({2}^{n + 1}\right) = \frac{1}{2}$

As,

$L < 1$, the series converges by the ratio test.