Finding a limit involving ratios of logarithms

Question

My friends. I apologize upfront for any spelling mistakes, I'm not used to writing math in english! I'm having trouble figuring out this limit and no calculator online I checked can show me (reasonable) steps for this.

The question tells me I have to use the ratio test. The limit I decided to call "L" is returning me "e", when it should return an "0". Can anyone point me to where I'm making it wrong?

Thank you very much in advance! Cheers.

$$\sum_{n=1}^{\infty}\dfrac{(-1)^n}{(\ln n)^n}\to\mathrm{Ratio\;test}\to\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\left|\dfrac{[\ln(n)]^n}{[\ln (n+1)]^{n+1}}\right|=\lim_{n\to\infty}\left[\dfrac{\ln(n)}{\ln(n+1)}\right]^n\dfrac{1}{\ln(n+1)}$$ $\displaystyle{\lim_{n\to\infty} f\cdot g=\lim_{n\to\infty} f\lim_{n\to\infty} g}$, $$\require{cancel}\lim_{n\to\infty}\left[\dfrac{\ln(n)}{\ln(n+1)}\right]^n\dfrac{1}{\ln(n+1)}=\lim_{n\to\infty}\left[\dfrac{\ln(n)}{\ln(n+1)}\right]^n\cancel{\lim_{n\to\infty}\dfrac{1}{\ln(n+1)}}$$ If the limit on the left exists, we are good to go: $$\lim_{n\to\infty}\left[\dfrac{\ln(n)}{\ln(n+1)}\right]^n=L$$ $$\ln(L)=\ln\lim_{n\to\infty}\left[\dfrac{\ln(n)}{\ln(n+1)}\right]^n\to\mathrm{L'Hospital}\to\ln(L)=\ln\lim_{n\to\infty}\left[\dfrac{1/n}{1/(n+1)}\right]^n\to\\\ln(L)=\ln\lim_{n\to\infty}\left[1+\dfrac1n\right]^n\to\ln(L)=\ln e\to \boxed{L=e}$$

Answer 1

Why not use the $\;n\,-$th root test better?:

$$\sqrt[n]{\left|\frac{(-1)^n}{\log^nn}\right|}=\frac1{\log n}\xrightarrow[n\to\infty]{}0$$

I can't understand in your work why you "erased" that $\;\frac1{\log(n+1)}\;$

Forcing the ratio test: you already had

$$\left(\frac{\log n}{\log(n+1)}\right)^n\cdot\frac1{\log(n+1)}$$

The rightmost factor clearly tends to zero, whereas the left one is bounded by one since the function $\;\log(x)\;$ is monotone ascending...

Answer 2

You have an error in applying L'Hôpital's Rule in taking the limit.

As I pointed out in a Comment earlier, you are taking a limit of a quantity which is less than $1$:

$$ L = \lim_{n\to\infty} \left[ \frac{\ln(n)}{\ln(n+1)} \right]^n $$

We cannot directly apply L'Hôpital's Rule to a subexpression as you tried to do with the ratio inside the brackets. The limit involves raising that quantity to a power. You applied the derivative to both numerator and denominator inside the bracket and got a ratio which is greater than $1$, so it clearly is invalid.

One idea to correct this would be to apply the power $n$ to both numerator and denominator before taking a derivative, e.g.

$$ L = \lim_{n\to\infty} \left[ \frac{(\ln(n))^n}{(\ln(n+1))^n} \right] $$

This doesn't seem like a great idea to pursue (too many tedious details to find the derivative with respect to $n$ when $n$ appears as an exponent).

Instead I would simply note that the ratio we've been discussing is bounded above by $1$. So any limit it may have is less than or equal to $1$.

In fact you can go back to the point where you removed the factor $\frac{1}{\ln(n+1)}$ and directly argue that because:

$$ 0 \lt \left[ \frac{\ln(n)}{\ln(n+1)} \right]^n \frac{1}{\ln(n+1)} \lt \frac{1}{\ln(n+1)} $$

the Pinching Lemma tells us the limit of the expression in between $0$ and $\frac{1}{\ln(n+1)}$ must be zero (because it is squeezed between two expressions both tending to zero).