Question

converge? diverge? \[\frac{ 1 }{ 2^n -n }\]

Answer 1

n goes to infinity?

Answer 2

oops sorry. n goes to 1

Answer 3

Can you write out the series? It's a little confusing...:\

Answer 4

\[\sum_{n=1}^{\infty}\frac{ 1 }{ 2^n -n }\]

Answer 5

here it is

Answer 6

Yay

Answer 7

i'm thinking you can use a Direct comparison test here, to compare \(\large a_{n}\) to \(\large b_{n}\)

Answer 8

\[\large a_{n} =\frac{1}{2^{n}-n} \ , \ b_{n}= \frac{1}{2^{n}}\] \(\large b_{n}\) converges because \(\large\frac12 < 1\)

Direct comparison test states that if \(\large a_{n} < b_{n} \ \text{AND} \ \ \sum{b_{n}} = C \) then \(\large \sum{a_{n}} = c\)

We can test that \(\large \frac{1}{2^{n}-n} < \frac{1}{2^{n}}\) therefore

\[\large \sum_{n=1}^{\infty} \frac{1}{2^n -n} =C\] by Direct Comparison Test.