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

Question

jhannybean · Answer

n goes to infinity?

anonymous · Answer

oops sorry. n goes to 1

jhannybean · Answer

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

anonymous · Answer

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

anonymous · Answer

here it is

jhannybean · Answer

Yay

jhannybean · Answer

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

jhannybean · Answer

$$\large a_{n} =\frac{1}{2^{n}-n} \ , \ b_{n}= \frac{1}{2^{n}}$$ $\large b_{n}$ 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.