Finding the interval of convergence for a series with lnn

In summary, the given series $\sum_{n = 2}^{\infty} \frac{1}{n\ln{(n)}}$ is divergent. One way to prove this is by using the integral test and showing that the integral of the function $\frac{1}{x\ln{(x)}}$ from 2 to infinity diverges, thus showing that the sum is also infinite.
  • #1
tmt1
234
0
So I have

$$\sum_{n = 2}^{\infty} \frac{1}{nln(n)}$$

I'm trying to apply the limit comparison test, so I can compare it to $b_n$ or $\frac{1}{n}$ and I can let $a_n = \frac{1}{nln(n)}$

Then I get $$\lim_{{n}\to{\infty}} \frac{n}{nln(n)}$$

Or $$\lim_{{n}\to{\infty}} \frac{1}{ln(n)}$$ Which is clearly 0.

Now, since the limit is 0, then the sums of series of $\sum_{ }^{}a_n$ would only converge if $\sum_{}^{}b_n$ converges. However $\sum_{}^{}b_n$ equals $\sum_{}^{}\frac{1}{n}$ which diverges.

The answer is that it is divergent, but I'm not sure how to prove it with my method. Is the method I am using not the right one to use, or am I misunderstanding the method?
 
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  • #2
tmt said:
So I have

$$\sum_{n = 2}^{\infty} \frac{1}{nln(n)}$$

I'm trying to apply the limit comparison test, so I can compare it to $b_n$ or $\frac{1}{n}$ and I can let $a_n = \frac{1}{nln(n)}$

Then I get $$\lim_{{n}\to{\infty}} \frac{n}{nln(n)}$$

Or $$\lim_{{n}\to{\infty}} \frac{1}{ln(n)}$$ Which is clearly 0.

Now, since the limit is 0, then the sums of series of $\sum_{ }^{}a_n$ would only converge if $\sum_{}^{}b_n$ converges. However $\sum_{}^{}b_n$ equals $\sum_{}^{}\frac{1}{n}$ which diverges.

The answer is that it is divergent, but I'm not sure how to prove it with my method. Is the method I am using not the right one to use, or am I misunderstanding the method?

I would use the integral test.

First let $\displaystyle \begin{align*} f(x) = \frac{1}{x\ln{(x)}} \end{align*}$. We should note that $\displaystyle \begin{align*} f'(x) = -\frac{\left[ \ln{(x)} + 1 \right] }{\left[ x \ln{(x)} \right] ^2 } \end{align*}$ which is negative for all $\displaystyle \begin{align*} x \geq 2 \end{align*}$, and thus $\displaystyle \begin{align*} \frac{1}{x\ln{(x)}} \end{align*}$ is decreasing in this region.

As it's decreasing, that means $\displaystyle \begin{align*} \sum_{n = 2}^{\infty} \frac{1}{n\ln{(n)}} > \int_2^{\infty}{ \frac{1}{x\ln{(x)}}\,\mathrm{d}x } \end{align*}$, so evaluating the integral we have

$\displaystyle \begin{align*} \int_2^{\infty}{ \frac{1}{x\ln{(x)}}\,\mathrm{d}x } &= \int_{\ln{(2)}}^{\infty}{ \frac{1}{u}\,\mathrm{d}u } \textrm{ after substituting } u = \ln{(x)} \implies \mathrm{d}u = \frac{1}{x}\,\mathrm{d}x \\ &= \lim_{\epsilon \to \infty} \left[ \ln{|u|} \right] _{\ln{(2)}}^{\epsilon} \\ &= \lim_{\epsilon \to \infty} \ln{(\epsilon )} - \ln{\left[ \ln{(2)}\right] } \\ &\to \infty \end{align*}$

As the sum is greater than this infinite value, the sum also diverges.
 

Related to Finding the interval of convergence for a series with lnn

1. What is the interval of convergence for a series with lnn?

The interval of convergence for a series with lnn is (0, ∞), meaning that the series will only converge for values of x that are greater than 0 and less than infinity.

2. How do you determine the interval of convergence for a series with lnn?

To determine the interval of convergence for a series with lnn, you can use the ratio test or the root test. These tests involve taking the limit of the absolute value of the terms in the series. If the limit is less than 1, the series will converge, and if it is greater than 1, the series will diverge.

3. Can the interval of convergence for a series with lnn include negative values?

No, the interval of convergence for a series with lnn cannot include negative values. This is because the natural logarithm function is only defined for positive values, and the series will diverge for negative values of x.

4. Are there any other methods for finding the interval of convergence for a series with lnn?

Yes, there are other methods for finding the interval of convergence for a series with lnn, such as the integral test and the comparison test. These methods involve comparing the series to other known convergent or divergent series.

5. What is the significance of the interval of convergence for a series with lnn?

The interval of convergence is important because it tells us for which values of x the series will converge. This information is useful in determining the behavior and properties of the series, such as its convergence or divergence, and the value it converges to.

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