# I Convergence of (ln n)^x

1. Jun 12, 2016

### Happiness

Find the values of $x$ for which the following series is convergent.

I compared the series with the harmonic series and deduced it is always divergent. I used $y^p<e^y$ for large $y$. I used a different method from the answer given, which I don't understand.

When $k=1$, $M_1=e=2.72$ and $M_0=1$. Since $M_0+1\leq n\leq M_1$, we have $n=2$, giving us the term $\frac{1}{1^X}$.

When $k=2$, $M_2=e^2=7.39$ and $M_1=e=2.72$. Since $M_1+1\leq n\leq M_2$, we have $n=4, 5, 6, 7$, giving us the terms $\frac{1}{2^X}+\frac{1}{2^X}+\frac{1}{2^X}+\frac{1}{2^X}$.

Finding the possible values of $n$ is troublesome.

Is there a typo in the given answer?

Last edited: Jun 12, 2016
2. Jun 12, 2016

### Staff: Mentor

You get of the order of e terms of 1/1X, e2 terms of 1/2X, e3 terms of 1/3X and so on (up to the constant prefactor). Following the same logic as your approximation, $\displaystyle \frac{e^n}{n^X}$ grows to infinity instead of going to zero.

3. Jun 12, 2016

### Happiness

Why is the series in the question and the series in the answer equivalent?

Their first terms are clearly different. The former's is $\frac{1}{(\ln 2)^X}$ while the latter's is $\frac{1}{1^X}$.

If they are not equivalent, how do we show that the former is always bigger than the latter for every term?

4. Jun 12, 2016

### Staff: Mentor

The first fraction is always larger than the second by construction of the Mk. So your series is larger than a divergent series.

5. Jun 12, 2016

### Happiness

Could you explain how? I don't see it.

6. Jun 12, 2016

### Staff: Mentor

ln(2) < 1
ln(3) < 2, ln(4) < 2, ln(5) < 2, ln(6) < 2, ln(7) < 2

Therefore 1/ln(2)X > 1/1X and so on for positive X.

7. Jun 12, 2016

### Happiness

The given answer is so complicated vs if we just compare the series with the harmonic series.

8. Jun 12, 2016

### Staff: Mentor

How do you do that for x=-3?

9. Jun 13, 2016

### Happiness

Let $y=\ln n$. For large $n, y^3<e^y$. So $(\ln n)^3<n$. Then $\frac{1}{(\ln n)^3}>\frac{1}{n}$. The series is bigger than the harmonic series for every term. Hence it is also divergent.

10. Jun 13, 2016

### Staff: Mentor

Ah right, that works as well, and it is easier.