Divergence of a series

**shinn** · March 15th 2010, 03:53 AM

Hi, I want to show that $\sum_{n=1}^{\infty}b_n$ , where $b_n = \ln(1 - \frac{1}{2^n})$ , is divergent (approaches negative infinity).

How would I go about approach this? I was thinking of using direct comparison or limit form of comparison test. However, I am not sure of what to compare $b_n$ with.

Any help will be kindly appreciated.

**Prove It** · March 15th 2010, 04:27 AM

Originally Posted by shinn

Hi, I want to show that $\sum_{n=0}^{\infty}b_n$ , where $b_n = \ln(1 - \frac{1}{2^n})$ , is divergent (approaches negative infinity).

How would I go about approach this? I was thinking of using direct comparison or limit form of comparison test. However, I am not sure of what to compare $b_n$ with.

Any help will be kindly appreciated.

It should be obvious that the very first term $\to \ln{0} \to -\infty$ .

If you can show the rest of the terms create a convergent series, when you add them, the sum is still going to $\to -\infty$ .

**shinn** · March 15th 2010, 04:50 AM

sorry I made a mistake when typing up the question, n is suppose to start at 1. thanks for pointing that out.

**Prove It** · March 15th 2010, 05:07 AM

$\sum_{n = 1}^{\infty}\ln{\left(1 - \frac{1}{2^n}\right)} = \ln{\left(1 - \frac{1}{2}\right)} + \ln{\left(1 - \frac{1}{4}\right)} + \ln{\left(1 - \frac{1}{8}\right)} + \ln{\left(1 - \frac{1}{16}\right)} + \dots$

$= \ln{\frac{1}{2}} + \ln{\frac{3}{4}} + \ln{\frac{7}{8}} + \ln{\frac{15}{16}} + \dots$

$= \ln{\left(\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{7}{8} \cdot \frac{15}{16} \cdot \dots\right)}$

$\to \ln{0}$

$\to -\infty$ .

**Krizalid** · March 15th 2010, 05:24 AM

the series converges.

**Prove It** · March 15th 2010, 05:25 AM

Originally Posted by Krizalid

the series converges.

Sorry, but I've just shown that it diverges...

**Krizalid** · March 15th 2010, 05:30 AM

Originally Posted by Prove It

$= \ln{\left(\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{7}{8} \cdot \frac{15}{16} \cdot \dots\right)}$

$\to \ln{0}$

$\to -\infty$ .

i don't get this step when you say $\to\ln0.$

mind if you explain it?

**Miss** · March 15th 2010, 05:31 AM

No. It converges.
Here is a hint:
Make a common denominator.
And use one of the logarithmic functions' properties.

**Prove It** · March 15th 2010, 05:39 AM

Originally Posted by Krizalid

i don't get this step when you say $\to\ln0.$

mind if you explain it?

When you multiply any number by a number $<1$ , the product is smaller than the original number.

Here I have continually multiplied by numbers $<1$ , so the product $\to 0$ .

**Prove It** · March 15th 2010, 05:44 AM

Originally Posted by Miss

No. It converges.
Here is a hint:
Make a common denominator.
And use one of the logarithmic functions' properties.

Sorry but I don't see how

$\ln{\left(1 - \frac{1}{2^n}\right)} = \ln{\left(\frac{2^n - 1}{2^n}\right)} = \ln{(2^n - 1)} - \ln{(2^n)}$

helps us in this case.

When you take the sum you just end up with

$\ln{1} - \ln{2} + \ln{3} - \ln{4} + \ln{7} - \ln{8} + \dots$ .

This is not a telescopic series so I don't see how you get that the series is convergent...

**Miss** · March 15th 2010, 05:52 AM

Originally Posted by Prove It

Sorry but I don't see how

$\ln{\left(1 - \frac{1}{2^n}\right)} = \ln{\left(\frac{2^n - 1}{2^n}\right)} = \ln{(2^n - 1)} - \ln{(2^n)}$

helps us in this case.

When you take the sum you just end up with

$\ln{1} - \ln{2} + \ln{3} - \ln{4} + \ln{7} - \ln{8} + \dots$ .

This is not a telescopic series so I don't see how you get that the series is convergent...

But I did not say it will be a telescoping series.

**Krizalid** · March 15th 2010, 05:54 AM

alright, so suppose i have the series $\sum{\ln \left( 1+\frac{1}{n^{2}} \right)}=\ln 2+\ln \frac{5}{4}+\ln +\frac{10}{9}+\cdots$ and i'm seeing here that the numerator grows faster than the denominator, thus the series diverges.

that's just wrong, doesn't provide a solid proof of the convergence of the series.

my series converges by limit comparison test with $b_n=\frac1{n^2}$ which is a solid proof of the convergence of the series.

**Prove It** · March 15th 2010, 05:59 AM

Considering that the OP has stated that he/she needs help getting to the answer of $-\infty$ he/she has been given, I have shown correctly how to get that answer, as the logarithm laws and logic I have used is not flawed.

**Prove It** · March 15th 2010, 06:00 AM

Originally Posted by Krizalid

alright, so suppose i have the series $\sum{\ln \left( 1+\frac{1}{n^{2}} \right)}=\ln 2+\ln \frac{5}{4}+\ln +\frac{10}{9}+\cdots$ and i'm seeing here that the numerator grows faster than the denominator, thus the series diverges.

that's just wrong, doesn't provide a solid proof of the convergence of the series.

my series converges by limit comparison test with $b_n=\frac1{n^2}$ which is a solid proof of the convergence of the series.

Actually it does work in this case.

Simple application of the same logarithm rule $\ln{a} + \ln{b} = \ln{(ab)}$ .

So $\ln{2} + \ln{\frac{5}{4}} + \ln{\frac{10}{9}} + \dots = \ln{\left(2 \cdot \frac{5}{4} \cdot \frac{10}{9} \cdot \dots \right)}$

and as stated before, since you're multiplying each term in the product by a number $>1$ , the product grows without bound (albeit slowly), as does the logarithm. So this series diverges.

**Krizalid** · March 15th 2010, 06:06 AM

$a_n=\ln\bigg(1-\frac1{2^n}\bigg)$ is strictly negative, so there's no difference on studying the series $\sum\ln\bigg(1-\frac1{2^n}\bigg)$ or $-\sum \ln\bigg(1-\frac1{2^n}\bigg)$ where the first one is negative and the second one positive, so limit comparison test with $\frac1{2^n}$ does apply, and the series converges.

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