Up and down, up and down

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Observations

• Black terms belong to the indices $k\in I_{n,0}$ and red terms belong to the indices $\red{k\in I_{n,1}}$ . Both sets contain $2^{n-1}$ indices
• Consecutive black and red terms cancel each other out: $\displaystyle\sum_{k\in I_n}a_k=\sum_{k\in I_n}b_k=0$ for all $n\geq 1$
• $\displaystyle\sum_{k\in I_{n,l}}a_k=\frac{(-1)^l}{2},\qquad \sum_{k\in I_{n,l}}b_k=\frac{(-1)^l}{2n}$

Proof: (last statement only) $\begin{aligned} \sum_{k\in I_{n,l}}a_k&=\frac{(-1)^l}{2^n}\cdot 2^{n-1}=\frac{(-1)^l}{2},&&&&\sum_{k\in I_{n,l}}b_k&=\frac{(-1)^l}{n2^n}\cdot 2^{n-1}=\frac{(-1)^l}{2n} \end{aligned}$

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Convergence

Let $A_N:=\sum_{k=2}^Na_k,\:B_N:=\sum_{k=2}^n b_k$ . Check $A_N$ first: $\begin{aligned} N&=2^{m}-1:&&&A_N&=\sum_{k=2}^Na_k=\sum_{n=1}^{m-1}\cancel{\sum_{k\in I_{n}}a_k}=0\\ N&=2^{m}+2^{m-1}-1:&&&A_N&=\sum_{k=2}^Na_k=\sum_{n=1}^{m-1}\cancel{\sum_{k\in I_{n}}a_k}+\sum_{k\in I_{m,0}}a_k=\frac{1}{2} \end{aligned}$

The sequence $A_N$ has two subsecquences $A_{2^m-1}=0,\:A_{2^m+2^{m-1}-1}=\frac{1}{2}$ with different limits - it cannot converge!

Let's tackle the other series with the Squeeze Theorem , take any $N\in I_m$ and calculate an upper estimate using the triangle inequality (*): $\begin{aligned} 0\leq |B_N|=\left|\sum_{k=2}^Nb_k\right |=\left|\sum_{n=1}^m\cancel{\sum_{k\in I_n}b_k}+\sum_{k=2^m}^Nb_k\right|\underset{(*)}{\leq}\sum_{k=2^m}^N|b_k|\leq\sum_{k\in I_m}|b_k|=\frac{1}{m2^m}\cdot 2^m=\frac{1}{m}\rightarrow0&&\text{for}&&m\rightarrow\infty \end{aligned}$

Using the Squeeze Theorem, $\lim_{N\rightarrow\infty}|B_N|=0$ . It follows that $\lim_{N\rightarrow\infty} B_N=0$ .

We proved $\boxed{\sum_{k=2}^\infty a_k\text{ diverges,}\qquad\sum_{k=2}^\infty b_k\text{ converges}}$

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Notes and common mistakes

• One might be tempted to argue "All terms cancel out, then the sum must be zero" . However, that argument is only true if we sum up the first $2^n$ elements - if we sum up to any point between two such borders, the sum might still take on any value!
• Checking only the sum of the first $2^n$ elements is the same as looking at only one subsequence of the sum. That is not enough to prove convergence!
• For both sequences, all standard convergence tests fail or are inconclusive. In particular, we cannot apply the alternating series test because $a_k,\:b_k$ do not change sign every term! We must check directly if $A_N,\:B_N$ converge.