Question

can someone please explain why the following series converges by comparison by 1/2^n

Answer 1

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

Answer 2

hint:
There is a theorem which states:
"given two positive terms series, say \[\sum {{a_n}} ,\;\sum {{b_n}} \] then if the subsequent condition holds:
\[\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = L \ne 0\]
then, those series are both convergent series or both divergent series

Answer 3

@Michele_Laino thanks. I understand the theorem, but I don't understand why the series should be compared to 1/2^n for limit comparison test?

Answer 4

and if an= original function and bn= 1/2^n, how do I tell if 1/2^n converge or diverge?

Answer 5

we have to evaluate this limit:
\[\large \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{{{2^n}}}}}{{\frac{1}{{{2^n} - 1}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{2^n} - 1}}{{{2^n}}} = \mathop {\lim }\limits_{n \to \infty } \left( {1 - \frac{1}{{{2^n}}}} \right) = ...?\]

Answer 6

i thought \[a _{n} = \frac{ 1 }{ 2^{n}-1 }\] ?

Answer 7

it is the same:
\[\large \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{{{2^n} - 1}}}}{{\frac{1}{{{2^n}}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{2^n}}}{{{2^n} - 1}} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{1 - \frac{1}{{{2^n}}}}}} \right) = ...?\]

Answer 8

that limit value it is equal to 1 in both cases

Answer 9

so, both series are convergent, since 1/2^n is convergent

Answer 10

namely, the series:
\[\sum {{b_n}} = \sum {\frac{1}{{{2^n} - 1}}} \]
is convergent too

Answer 11

how did you determine that 1/2^n is convergent?

Answer 12

@Michele_Laino

Answer 13

it is very simple: The general formula for a sum S_n of a geometric series, is:
\[{S_n} = \frac{{1 - {a^{n + 1}}}}{{1 - a}}\]
where a=1/2, in our case. Please, keep in mind that we can demonstrate that formula, in many ways, for example using the Mathematical Induction Principle

Answer 14

So substituting a=1/2, we get:
\[{S_n} = \frac{{1 - {{\left( {\frac{1}{2}} \right)}^{n + 1}}}}{{1 - \frac{1}{2}}}\]

Answer 15

then we compute the limit value of that expression as n--->infinity, so we get:
\[S = {S_\infty } = 2\]
then we can write:
\[\Large \sum {\left( {\frac{1}{{{2^n}}}} \right)} = 2\]

Answer 16

so this is like a regular geometric series where r=1/2? why did you think of using a geometric series? how did you think that out?

Answer 17

we have a geometric series whe the ratio between one term and its previous term is constant for all terms of the series. In our case the ratio between one term and its previous term is:
\[\Large \frac{{\frac{1}{{{2^n}}}}}{{\frac{1}{{{2^{n - 1}}}}}} = \frac{{{2^{n - 1}}}}{{{2^n}}} = \frac{1}{2}\]

Answer 18

ah i see. thank you so much!

Answer 19

thank you!