# Thread: Convergence / Divergence

1. ## Convergence / Divergence

Question: Determine whether the series is convergent or divergent, and if it is possible, find the sum.

$\sum_{n = 1}\ln(\frac{n}{2n + 5})$

At this point in my class, we only know geometric series, the divergence test, and the p-series. So we cant use integral test, comparison test etc. My initial attempt was to use the divergence test. But I am having a hard time evaluating the limit. Thanks in advance.

p.s the sigma goes from n=1 to infinity.

2. Originally Posted by zellster87
Question: Determine whether the series is convergent or divergent, and if it is possible, find the sum.

$\sum_{n = 1}\ln(\frac{n}{2n + 5})$

At this point in my class, we only know geometric series, the divergence test, and the p-series. So we cant use integral test, comparison test etc. My initial attempt was to use the divergence test. But I am having a hard time evaluating the limit. Thanks in advance.

p.s the sigma goes from n=1 to infinity.
the natural log is a "nice" continuous function, therefore we can pass the limit "through" it. so $\lim_{n \to \infty} \ln \left( \frac n{2n + 5} \right) = \ln \left( \lim_{n \to \infty} \frac n{2n + 5}\right)$

can you continue?

3. I think so.

$
\lim_{n \to \infty} \ln \left( \frac n{2n + 5} \right) = \ln \left( \lim_{n \to \infty} \frac n{2n + 5}\right)
$

is equal to:

$
\ln \left( \lim_{n \to \infty}\frac{1}{2 + \frac{5}{n}}\right)
$

Then from that, i assume the limit is equal to 1/2?

Would I then have to take the natural log on 1/2, or is it automatically divergent since the limit is not equal to zero.

Thanks for you reply.

4. Originally Posted by zellster87
I think so.

$
\lim_{n \to \infty} \ln \left( \frac n{2n + 5} \right) = \ln \left( \lim_{n \to \infty} \frac n{2n + 5}\right)
$

is equal to:

$
\ln \left( \lim_{n \to \infty}\frac{1}{2 + \frac{5}{n}}\right)
$

Then from that, i assume the limit is equal to 1/2?

Would I then have to take the natural log on 1/2, or is it automatically divergent since the limit is not equal to zero.

Thanks for you reply.
yes, you are correct. ln(1/2) is not zero, we can stop there. the series diverges by the test for divergence

5. Got it. Thanks for your help.

6. Originally Posted by zellster87
Question: Determine whether the series is convergent or divergent, and if it is possible, find the sum.

$\sum_{n = 1}\ln(\frac{n}{2n + 5})$

At this point in my class, we only know geometric series, the divergence test, and the p-series. So we cant use integral test, comparison test etc. My initial attempt was to use the divergence test. But I am having a hard time evaluating the limit. Thanks in advance.

p.s the sigma goes from n=1 to infinity.
As $n \to \infty,\ \ln\left( \frac{n}{2n + 5} \right) \to \ln(1/2) \ne 0$ so the series diverges.

For a series to converge the limit of the sequence of terms must be 0.

RonL