# Show that the series $\sum_{n>0} \frac{(-1)^n}{n}\left \lfloor \frac{\ln n}{\ln 2} \right \rfloor$ converges

Show that the series $$\sum_{n>0} \frac{(-1)^n}{n}\left \lfloor \frac{\ln n}{\ln 2} \right \rfloor$$ converges, where $$\left \lfloor x \right \rfloor$$ is the integer part of $$x$$.

I have used the fact that

$$\ln n\rightarrow\sum_{k=1}^{n} \frac{1}{k}$$

but its difficult to express the integer part of this sum.

Using the hint in the answer, I need to show that :

1. $$a_n=\frac{1}{n} \left \lfloor \frac{ln(n)}{ln(2)} \right \rfloor$$ have a constant signe
2. $$a_n \rightarrow 0$$ and decreasing .

The first point is obvious because $$\frac{ln(n)}{n} \rightarrow 0$$

I have to show that $$a_{n+1} to apply the alternative series test i am stuck here

For this sum, Alternating Series Test is not directly applicable because the magnitudes of the summands do not decay monotonely:

For the proof, it is convenient to group terms according to the value of $$\lfloor \log_2 n\rfloor$$. Indeed, for each given $$N$$, we may write

$$\sum_{n=1}^{N} \frac{(-1)^n}{n} \lfloor \log_2 n \rfloor = \sum_{n=1}^{N} \sum_{m=1}^{\infty} \frac{(-1)^n}{n} \cdot m \mathbf{1}_{\{ 2^m \leq n < 2^{m+1} \}} = \sum_{m=1}^{\infty} A_{N,m}, \tag{1}$$

where $$A_{N,m}$$ is defined by

$$A_{N,m} = m \sum_{\substack{n \leq N \\ 2^m \leq n < 2^{m+1}}} \frac{(-1)^n}{n}.$$

Now we refer to the following intermediate observation in the proof of Alternating Series Test:

Lemma. Let $$(a_k)_{m \leq k \leq n}$$ be a sequence of non-negative and non-increasing real numbers. Then

$$\left| \sum_{m \leq k \leq n} (-1)^k a_k \right| \leq a_m.$$

From this, we immediately find that

$$|A_{N,m}| \leq \frac{m}{2^m} \quad \text{for all} \quad N, m \geq 1.$$

Since this bound is summable in $$m$$, meaning that $$\sum_{m\geq 1} m/2^m < \infty$$, Weierstrass $$M$$-test shows that the sum in $$\text{(1)}$$ converges as $$N\to\infty$$. Moreover, the limit can be computed by taking limit term-wise:

$$\lim_{N\to\infty} \sum_{n=1}^{N} \frac{(-1)^n}{n} \lfloor \log_2 n \rfloor = \sum_{m=1}^{\infty} \lim_{N\to\infty} A_{N,m}.$$

• Similar to mine, with a different way of bounding the sums. Sep 23 '19 at 6:14

Consider the terms from $$n=2^m$$ to $$2^{m+1}-1$$. There are $$2^m$$ of these, and for each the integer part is $$m$$.

Therefore the sum over these is

$$m\sum_{k=0}^{2^m-1}(-1)^k/(2^m+k)\\ =m\sum_{k=0}^{2^{m-1}-1}(1/(2^m+2k)-1/(2^m+2k+1))\\ =m\sum_{k=0}^{2^{m-1}-1}(1/((2^m+2k) (2^m+2k+1)))\\ \lt m\sum_{k=0}^{2^{m-1}-1}(1/((2^m) (2^m)))\\ =\dfrac{m}{2^{m+1}}$$

and the sum of these converges.

The series restricted to $$[2^m,2^{m+1}-1]$$ equals $$m\sum_{n=2^m}^{2^{m+1}-1}\frac{(-1)^n}{n}=m\left[\sum_{n=2^m}^{2^{m+1}-1}-\frac{1}{n}+\sum_{\substack{n\in[2^m,2^{m+1}-1]\\n\text{ even}}}\frac{2}{n}\right]$$ or $$m\left[-H_{2^{m+1}-1}+2H_{2^m-1}-H_{2^{m-1}-1}\right]= m\left[-H_{2^{m+1}}+2H_{2^m}-H_{2^{m-1}}+\frac{1}{2^{m+1}}-\frac{2}{2^m}+\frac{1}{2^{m-1}}\right].$$ Since $$H_N = \log N+\gamma +\frac{1}{2N}+O\left(\frac{1}{N^2}\right)$$ we have $$m\sum_{n=2^m}^{2^{m+1}-1}\frac{(-1)^n}{n} = m\left[\frac{1}{2^{m+2}}+O\left(\frac{1}{4^m}\right)\right]$$ and the original series is convergent by comparison with $$\sum_{m\geq 0}\frac{m}{2^{m+2}}$$.