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

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Answer 1 · 2017-01-08T07:05:54

The series converges by the ratio test.

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->oo)2^n/(2^(n+1))=1/2#

As,

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