Choosing $N$ so that the sum is greater than 10

Question

I have to find a number $N$ such that the sum

$$\sum_{n=3}^N \dfrac{1}{n(\log_2 n)(\log_2\log_2 n)} > 10.$$

I know that the series $$\sum_{n=3}^{\infty} \dfrac{1}{n(\log_2 n)(\log_2\log_2 n)}$$ diverges by using the Cauchy Condensation Test. However, it diverges very slowly. For example, when I take $N=100$, the sum is even less than 1.

So, I don't know how to find the number $N$ such that the sum is greater than 10.

Any help is highly appreciated.

Answer 1

Since your function is monotonically decreasing, we know that $$ \int_3^N\frac{1}{x\log_2(x)\log_2\log_2(x)}dx < \sum_{n=3}^N \frac{1}{x\log_2(x)\log_2\log_2(x)} $$ This integral can be evaluated $$ \int_3^N\frac{1}{x\log_2(x)\log_2\log_2(x)}dx = \log(2)^2\log_2(\log_2(\log_2(x)))|_3^N $$ We want to find $N$ such that $$ \log(2)^2\log_2(\log_2(\log_2(N)))-\log(2)^2\log_2(\log_2(\log_2(3)))>10 $$

And now from here you can play around with the inequality and algebra to get a bound for $N$

Answer 2

This may be overkill (you don't need to find the least $N$ do you?)

The harmonic series is know to diverge and we can find and $M$ so that $\sum_{j=1}^M \frac 1j > 10$. One such formula is $\sum_{j=1}^{2^k} \frac 1j > 1+\frac k2$ so $\sum_{j=1}^{2^{18}} \frac 1j > 1 +\frac{18}2=10$ but you can maybe find another formula.

Now $a_n= n\log_2 n \log_2\log_2 n$ is increasing but very slowly. I think it is a safe but and probably easy to show that $a_{k+1} -a_k < 1$.

Therefore there is a series of indexes, $k_1, k_2, k_3,......$ where $a_{k_1} < 1 <a_{k_2}$ and $a_{k_j} < j < a_{k_{j+1}}$

Example $a_3 = 3\log_2 3\log_2 \log_2 3 < 1$ so we can say $k_1 =3$ and if $m$ is such that $1 < m\log_2 m \log_2 \log_2 m < 2$ we can set $k_2 = m$ and so on.

Now $\sum_{n=3}^M \frac 1{a_n} < \sum_{n=k_i; i=1}^{k_W: k_W\le M; k_{W+1}>M} \frac 1{a_{k_i}}>\sum_{n=k_i;i=1}^{k_W} \frac 1i = \sum_{i=1}^W \frac 1i$.

So to solve $\sum_{n=3}^M \frac 1{a_n} > 10$ it suffices to solve for $W$ so $\sum_{i=1}^W\frac 1i > 10$. We can solve that with $W = 2^{18}$

Then we need find a $k_W$ so that $2^{18} -1 < k_W \log_2 k_W \log_2 \log_2 k_W < 2^{18}$. And that can be our value of $M$.

===

Okay, overkill.

For $n \ge 3$ we know $0< \log_2 \log_2 n < \log_2 n < n$ so $n\log_2 n \log_2 \log_2 n > (\log_2 \log_2 n)^3$ and $(\log_2 \log_2 n)^3\ge 2^{18}$ if $\log_2 \log_2 n \ge 2^6$ which is true if $\log_2 n \ge 2^{2^6}$ which is true if $n \ge 2^{2^{2^6}}$.

So consider $\sum_{k=3}^{2^{2^{2^6}}} \frac 1{k\log_2 k \log_2 \log_2 k} > \sum_{k=3}^{2^{2^{2^6}}} \frac 1{(\log_2 \log_2 k)^3} > \sum_{k=3}^{2^{2^{2^6}}} \frac 1{\lceil (\log_2\log_2 k)^3\rceil} \le$
$\sum_{k\in A} \frac 1{\lceil (\log_2\log_2 k)^3\rceil}$

We $A$ is any subset of $\{3,...., 2^{2^{2^6}}\}$

Now we can find a subset of $\{3,...., 2^{2^{2^6}}\}$ so that $A= \{a_1,a_2, ....., a_{2^{18}}\}$ where $\lceil (\log_2\log_2 a_j)^3 \rceil = j$ (Let's put a pin in that for now)

So $\sum_{k=3}^{2^{2^{2^6}}} \frac 1{k\log_2 k \log_2 \log_2 k} >$
$\sum_{a_j\in A}\frac 1{\lceil (\log_2\log_2 a_j)^3 \rceil} =\sum_{a_j\in A}\frac 1{j}=$
$\sum_{j=1}^{2^{18}} \frac 1j \ge 1 + \frac {18}2 = 10$.

Ta-da.....

All that remains is our pin. $0 =\log_2 \log_2 2 < \log_2\log_2 3 < \log_2\log_2 4=1$ so $0 < (\log_2\log_2 3)^3 < 1$ and so $\lceil (\log_2\log_2 3)^3 \rceil = 1$ and so let $a_1 = 3$.

Now $(\log_2 \log_2 (n+1))^3 - (\log_2\log_2 n)^3 < 1$ (another pin) so for every $K$ there is an $n$ so that $K-1 < (\log_2\log_2 n)^3 \le K$ so we can in such a case set $a_K = $ that $n$. And as $(\log_2 \log_2 2^{2^{2^6}})^3 = 2^{18}$ we can set $a_{2^{18}} = 2^{2^{2^6}}$.

... guess the only thing left is to prove $(\log_2 \log_2 (n+1))^3 - (\log_2\log_2 n)^3 < 1$ for all $n$....