How do I evaluate this limit without l'Hôpital’s rule?

Question

Context:

I recently had this question given to me:

Does the following series converge or diverge $$\sum_{n=3}^{\infty} \frac{1}{n \ln n \sqrt{\ln(\ln n)}}$$

My working:

As the sum is decreasing and nonnegative we can use the Cauchy condensation test. Applying it once we obtain, $$\sum_{n=3}^{\infty} \frac{1}{n \ln n \sqrt{\ln(\ln n)}} \to \sum_{n=3}^{\infty} \frac{2^n}{2^n \ln 2^n \sqrt{\ln \left(\ln 2^n \right)}} \to \frac{1}{\ln 2} \sum_{n=3}^{\infty} \frac{1}{n \sqrt{\ln \left( n \ln 2 \right)}}$$ Applying the condensation test once more we have, $$\frac{1}{\ln 2} \sum_{n=3}^{\infty} \frac{1}{n \sqrt{\ln \left( n \ln 2 \right)}} \to \frac{1}{\ln 2} \sum_{n=3}^{\infty} \frac{2^n}{2^n \sqrt{\ln \left( 2^n \ln 2 \right)}} \to \frac{1}{\ln 2} \sum_{n=3}^{\infty} \frac{1}{ \sqrt{n \ln 2 + \ln 2 + \ln \ln 2}}$$

Again, $$\frac{1}{\ln 2} \sum_{n=3}^{\infty} \frac{1}{\sqrt{\ln n + \ln 2 + \ln \ln 2}} \to \frac{1}{\ln 2} \sum_{n=3}^{\infty} \frac{2^n}{\sqrt{\ln 2^n + \ln 2 + \ln \ln 2}} \to \frac{1}{\ln 2} \sum_{n=3}^{\infty} \frac{2^n}{\sqrt{n \ln 2 + \ln 2 + \ln \ln 2}}$$ Now with a factor of $2^{n}$ in the numerator, the sequence in the above series no longer tends to 0 and so the series would diverge by the divergence test.

My tutor's feedback:

He stated that it wasn't enough to introduce a factor of $2^n$ and state that it would diverge by the divergence test, I needed to show that the series diverged by actually using the divergence test.

So I needed to evaluate the limit, $$\lim_{n \to \infty} \frac{2^n}{\sqrt{n \ln 2 + \ln 2 + \ln \ln 2}}$$ It is easy to evaluate using l'hopital’s rule. We see the the limit is $\infty$ which is greater than $1$, so the series diverges.

My Question

Is there a way to evaluate this without using l'hopital’s rule? Please do share.

Answer 1

Based on the well-known inequality $e^x≥x+1$, we have :

$$ \begin{align}2^n=e^{\ln 2^n}&≥\ln 2^n+1\\ &=n\ln 2+1\end{align} $$

Then, you can observe that :

$$ \begin{align}f(n)&=\frac{2^n}{\sqrt{n \ln 2 + \ln 2 + \ln \ln 2}}\\ &>\frac {2^n}{\sqrt {n\cdot 1+n+n}}\\ &=\frac {\sqrt 3}{3}\cdot \frac {2^n}{\sqrt n}\\ &>\frac {\sqrt 3}{3}\cdot \frac {n\ln 2+1}{\sqrt n}\\ &=\frac {\sqrt 3}{3}\cdot \left (\sqrt n\ln 2+\frac {1}{\sqrt n}\right)\\ &>\underbrace{\frac {\sqrt 3\ln 2}{3}}_{>0}\cdot \sqrt n,\thinspace\thinspace\thinspace \forall n≥1\thinspace .\end{align} $$

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$\rm {Another\thinspace\thinspace variation :}$

Since,

$$ \begin{align}2^n&≥n\ln 2+1>n\ln 2\end{align} $$

we can also obtain the same result as follows :

$$ \begin{align}f(n)&=\frac{2^n}{\sqrt{n \ln 2 + \ln 2 + \ln \ln 2}}\\ &>\frac{n\ln 2}{\sqrt{n \ln 2 + \ln 2 + \ln \ln 2}}\\ &>\frac {n\ln 2}{\sqrt {n\ln 2+n\ln 2+n\ln 2}}\\ &=\underbrace{\frac {\sqrt {3\ln 2}}{3}}_{>0}\cdot \sqrt n, \thinspace\thinspace\thinspace \forall n≥1\thinspace .\end{align} $$

Answer 2

You can divide the numerator and denominator by $\sqrt{n}$. That way the denominator will go toward $\sqrt{\ln 2}$ and the numerator toward infinity:

Given: $$\lim_{n \to \infty} \frac{2^n}{\sqrt{n \ln 2 + \ln 2 + \ln \ln 2}}$$ Divide numerator and denominator by $\sqrt{n}$: $$\lim_{n \to \infty} \frac{\frac{2^n}{\sqrt{n}}}{\frac{\sqrt{n \ln 2 + \ln 2 + \ln \ln 2}}{\sqrt{n}}}$$ Use limit quotient property: $$\frac{\lim_{n \to \infty} \frac{2^n}{\sqrt{n}}}{\lim_{n \to \infty} \frac{\sqrt{n \ln 2 + \ln 2 + \ln \ln 2}}{\sqrt{n}}}$$ Evaluate numerator: $$\lim_{n \to \infty} \frac{2^n}{\sqrt{n}}=\infty$$ Evaluate demoninator: $$\lim_{n \to \infty} \frac{\sqrt{n \ln 2 + \ln 2 + \ln \ln 2}}{\sqrt{n}}=\sqrt{\ln 2}$$ Therefore: $$\lim_{n \to \infty} \frac{\frac{2^n}{\sqrt{n}}}{\frac{\sqrt{n \ln 2 + \ln 2 + \ln \ln 2}}{\sqrt{n}}}=\frac{\infty}{\sqrt{\ln 2}}=\infty$$

Answer 3

We can prove that $\lim_{n\to\infty}2^n / n = \infty$ directly from the definition of an infinite limit, and hence without any need of L'Hôpital's. I'm not sure if proofs working with the definition of a limit have been covered in your course, so I'll omit the proof.

Writing

\begin{align*} \frac{2^n}{\sqrt{n\ln2+\ln2+\ln\ln2}} &= \frac{\frac{2^n}{n}}{\frac{\sqrt{n\ln2+\ln2+\ln\ln2}}{n}}\\ &= \frac{\frac{2^n}{n}}{\frac{\sqrt{n}\sqrt{\ln2+\frac{\ln2}{n}+\frac{\ln\ln2}{n}}}{n}}\\ &= \frac{\frac{2^n}{n}}{\frac{\sqrt{\ln2+\frac{\ln2}{n}+\frac{\ln\ln2}{n}}}{\sqrt{n}}}\\ &= \sqrt{n}\cdot\frac{2^n}{n}\cdot\frac{1}{\sqrt{\ln2+\frac{\ln2}{n}+\frac{\ln\ln2}{n}}}\\ \end{align*}

and noting that $\sqrt{n}\to\infty$ and $\sqrt{\ln2+\frac{\ln2}{n}+\frac{\ln\ln2}{n}}\to\sqrt{\ln 2}$ as $n\to\infty$ immediately implies the desired limit:

\begin{align*} \lim_{n\to\infty}\frac{2^n}{\sqrt{n\ln2+\ln2+\ln\ln2}} &= \lim_{n\to\infty}\left(\sqrt{n}\cdot\frac{2^n}{n}\cdot\frac{1}{\sqrt{\ln2+\frac{\ln2}{n}+\frac{\ln\ln2}{n}}}\right)\\ &= \infty\cdot\infty\cdot\frac{1}{\sqrt{\ln 2}}\\ &=\infty \end{align*}

That said, I should mention that you made two mistakes in your work. The first of these is not major:

$$\dots=\frac{1}{\ln 2}\sum_{n=3}^\infty\frac{2^n}{2^n\sqrt{\ln(2^n\ln 2)}}=\frac{1}{\ln 2}\sum_{n=3}^\infty\frac{1}{\sqrt{n\ln2+\ln2+\ln\ln2}}$$

I presume you are saying that the terms of these two series are equivalent. This is not the case; we actually have

$$\frac{1}{\ln 2}\sum_{n=3}^\infty\frac{2^n}{2^n\sqrt{\ln(2^n\ln 2)}}=\frac{1}{\ln 2}\sum_{n=3}^\infty\frac{1}{\sqrt{n\ln2 + \ln\ln2}}$$

with no extra $\ln2$ in the denominator of the second sum.

The second mistake is a bit more concerning:

Again, $$\frac{1}{\ln2}\frac{1}{\sqrt{\ln n+\ln2 + \ln\ln2}}=\frac{1}{\ln 2}\sum_{n=3}^\infty\frac{2^n}{\sqrt{n\ln2+\ln2+\ln\ln2}}$$

I think you forgot to add a summation in the left. More importantly, I'm unsure where you got this equality. The left side has an $\ln n$ term in the denominator that is not present in your previous series, and you suddenly added a factor of $2^n$ in the numerator on the right. These were cancelled in previous work; where does this come from? Our final series should be

$$\frac{1}{\ln 2}\sum_{n=3}^\infty\frac{1}{\sqrt{n\ln2 + \ln\ln2}}$$

which is not amenable to the test for divergence because the terms tend to $0$.

Answer 4

Consider the sequence $x_n=\frac{\sqrt{n \ln 2 + \ln 2 + \ln \ln 2}}{n^2}$

$x_n\to 0$ as $n\to\infty$

Since $(x_n) $ is convergent, it is bounded i.e $\exists M>0$ such that $|x_n|<M$ for all $n\in\Bbb{N}$

Implies $\sqrt{n \ln 2 + \ln 2 + \ln \ln 2}<M{n^2}$ for all $n\in \Bbb{N}$

Hence $\frac{2^n}{\sqrt{n \ln 2 + \ln 2 + \ln \ln 2}}>\frac{2^n}{Mn^2}>\frac{n}{M}$ for all $n\ge 10$

Hence the sequence$(\frac{2^n}{\sqrt{n \ln 2 + \ln 2 + \ln \ln 2}}) $ is unbounded above.

To confirm it won't oscillate,see it's an increasing sequence.

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Quick way:

Let $y_n= \frac{2^n}{\sqrt{n \ln 2 + \ln 2 + \ln \ln 2}}$

$$\begin{align}\lim_{n\to\infty}\frac{y_{n+1}}{y_n}&=\lim_{n \to \infty} \frac{2^{n+1}({\sqrt{n \ln 2 + \ln 2 + \ln \ln 2})} }{2^n(\sqrt{(n+1) \ln 2 + \ln 2 + \ln \ln 2})}\\&=2>1\end{align}$$

Hence $(y_n) \to\infty$

Answer 5

Here is a slightly more direct approach:

$$\sum_n\frac{1}{n\log n (\log\log n)^\alpha}$$ converges iff $I:=\int^\infty_3\frac{1}{x\log x(\log\log x)^{\alpha}}\,dx $ converges. The substation $x=e^y$ yields $$I=\int^\infty_{\log 3}\frac{1}{y\log^{\alpha}(y)}\,dy=\left\{ \begin{array}{lcr} \infty &\text{if} & 0<\alpha\leq 1\\ \frac{\log^{\alpha-1}(\log 3)}{1-\alpha} &\text{if}& \alpha>1 \end{array} \right.$$

Answer 6

Your work is almost fine, indeed from this step

$$\frac{1}{\ln 2} \sum_{n=3}^{\infty} \frac{1}{n \sqrt{\ln \left( n \ln 2 \right)}} \to \frac{1}{\ln 2} \sum_{n=3}^{\infty} \frac{2^n}{2^n \sqrt{\ln \left( 2^n \ln 2 \right)}} \to \frac{1}{\ln 2} \sum_{n=3}^{\infty} \frac{1}{ \color{red}{\sqrt{n \ln 2 + \ln \ln 2}}}$$

we can conclude for divergenge by limit comparison test with $\sum_{n=3}^{\infty} \frac{1}{ \sqrt n}$.