We start by proving that $$ 1⟹<!--\; ⟹\; -->2$$ so assume that series converges. Let $$ S_n:=\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}{a}_k$$ be the sequence of partial sums, therefore we know that for any $$ \mathrm{ϵ<!--\; ϵ\; -->}$$ there is an $$ N\in <!--\; \in \; -->{\mathbb{N}}_0$$ such that for all $$ n\ge <!--\; \ge \; -->N,|S_n-<!--\; -\; -->L|<<!--\; <\; -->\mathrm{ϵ<!--\; ϵ\; -->}$$, note that then it follows that $\[ \left|\underset{k=n+1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{a}_k\right|=|S_n-<!--\; -\; -->L|<<!--\; <\; -->\mathrm{ϵ<!--\; ϵ\; -->} \]$ as needed.

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Now we'll prove that $$ 2⟹<!--\; ⟹\; -->3$$ so to show this is true, let $$ \mathrm{ϵ<!--\; ϵ\; -->}\in <!--\; \in \; -->{\mathbb{R}}^{+}$$ therefore by considering $$ \frac{\mathrm{ϵ<!--\; ϵ\; -->}}{2}$$ in our assumption we get an $$ N\in <!--\; \in \; -->{\mathbb{N}}_1$$ such that for any $$ n\ge <!--\; \ge \; -->N$$ we have that $\[ \left|\underset{k=n+1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{a}_k\right|\le <!--\; \le \; -->\frac{\mathrm{ϵ<!--\; ϵ\; -->}}{2} \]$

Now let $$ a,b\ge <!--\; \ge \; -->N$$ and we'll prove that $$ \left|\underset{k=a+1}{\overset{b}{\sum <!--\; \sum \; -->}}{a}_k\right|<<!--\; <\; -->\mathrm{ϵ<!--\; ϵ\; -->}$$, to do this note the following as needed.

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We'll prove that $$ 3⟹<!--\; ⟹\; -->1$$, since we assume $$ 3$$ and use the notation for $$ S_n$$ as the partial sums, then we can see that $\[ \left|\underset{k=n+1}{\overset{m}{\sum <!--\; \sum \; -->}}\right|=|S_m-<!--\; -\; -->S_n| \]$ and thus assuming $$ 3$$ is the same as assuming that $$ S_n$$ is cauchy, but we know that any cauchy sequence is convergent and thus $$ S_n$$ converges, which shows that $$ \underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{a}_n$$ converges.

Assume our hypothesis, and we'll show that $$ \left(a_n\right)$$ is summable, to do this we'll use 2 of the cauchy criterion, so let $$ \mathrm{ϵ<!--\; ϵ\; -->}\in <!--\; \in \; -->{\mathbb{R}}^{+}$$, also since $$ \left(b_n\right)$$ is summable, by $$ 3$$ of the cauchy criterion we obtain some $$ N\in <!--\; \in \; -->{\mathbb{N}}_1$$ such that for any $$ n,m\ge <!--\; \ge \; -->N$$ we have $$ \left|\underset{k=n+1}{\overset{m}{\sum <!--\; \sum \; -->}}{b}_k\right|<<!--\; <\; -->\mathrm{ϵ<!--\; ϵ\; -->}$$, but recall the inequalities we've assumed between $$ \left(a_n\right),\left(b_n\right)$$, which shows us that $\[ \left|\underset{k=n+1}{\overset{m}{\sum <!--\; \sum \; -->}}{a}_k\right|\le <!--\; \le \; -->\underset{n=k+1}{\overset{m}{\sum <!--\; \sum \; -->}}\left|a_k\right|\le <!--\; \le \; -->\underset{k=n+1}{\overset{m}{\sum <!--\; \sum \; -->}}{b}_k<<!--\; <\; -->\mathrm{ϵ<!--\; ϵ\; -->} \]$ The first $$ \le <!--\; \le \; -->$$ follows from the generalized triangle inequality, the second follows from our assumed inequality. Thus $$ \left(a_n\right)$$ is summable.

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Suppose that $$ \left(a_n\right)$$ is not summable, we'll show that $$ \left(b_n\right)$$ is not summble, for if it were then by part one of the proof then $$ \left(a_n\right)$$ would be which is a contradiction.

Let's define $$ S_n=\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}{a}_k$$ and $$ T_n=\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}{2}^k{a}_{2^k}$$.

Assume that $$ \underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{2}^n{a}_{2^n}$$ converges, which by definition means that $$ \left(T_n\right)$$ converges and is therefore bounded above, which means there is some $$ M$$ such that for any $$ i\in <!--\; \in \; -->{\mathbb{N}}_0,T_i\le <!--\; \le \; -->M$$, we'll use this fact to show that $$ \left(S_n\right)$$ is bounded above by showing that it has a subsequence which is bounded above. Namely we will prove that $\[ S_{2^n-<!--\; -\; -->1}\le <!--\; \le \; -->t_{n-<!--\; -\; -->1}\le <!--\; \le \; -->M \]$

Base case $$ n=1$$, $$ S_{2^1-<!--\; -\; -->1}=S_1=a_1=2^0{a}_{2^0}=T_0\le <!--\; \le \; -->M$$ as needed. Now asssume that it holds true for $$ n\in <!--\; \in \; -->{\mathbb{N}}_1$$, which is to say that $$ S_{2^n-<!--\; -\; -->1}\le <!--\; \le \; -->T_{n-<!--\; -\; -->1}\le <!--\; \le \; -->M$$ and now let's prove that $$ S_{2^{n+1}-<!--\; -\; -->1}\le <!--\; \le \; -->T_n\le <!--\; \le \; -->M$$

Note that in the induction step the following fact will be useful, which says for any $$ n\in <!--\; \in \; -->{\mathbb{N}}_1$$ we have that $\[ T_n-<!--\; -\; -->T_{n-<!--\; -\; -->1}=2^n{a}_{2^n} \]$ Now continuing on, we note the following

Note that going from the second line to the third comes from the fact that $$ \left(a_n\right)$$ is decreasing which means for any $$ x\in <!--\; \in \; -->\{2_n,\dots <!--\; \dots \; -->,2^{n+1}-<!--\; -\; -->1\}$$ that $$ a_{2^n}\ge <!--\; \ge \; -->a_x$$ additionally the the size of that set is $$ 2^{n+1}-<!--\; -\; -->1-<!--\; -\; -->2^n+1=2^n(2-<!--\; -\; -->1)=2^n$$, which allow us to upper bound that summation by $$ 2^n{a}_{2^n}$$, which shows the induction step true. Thus we've proven that $$ S_n$$ is bounded above, and so by the MCT it converges which is to say that $$ \underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{a}_n<<!--\; <\; -->\mathrm{\infty <!--\; \infty \; -->}$$

Now let's assume that $$ \underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{2}^n{a}_{2^n}$$ converges, and show that the other one does as well, we'll follow a similar structure that in we show that $$ T_n$$ is bounded above. Specifically we will show that $\[ a_1+2\underset{i=2}{\overset{2^n}{\sum <!--\; \sum \; -->}}{a}_n\ge <!--\; \ge \; -->T_n \]$ Since $$ \left(S_n\right)$$ is bounded above by some $$ M\in <!--\; \in \; -->\mathbb{R}$$ then by adding $$ a_1$$ to both sides we obtain $$ 2S_{2^n}\ge <!--\; \ge \; -->a_1+T_n$$ which is to say that $$ 2M\ge <!--\; \ge \; -->a_1+T_n$$ which shows that $$ T_n$$ is bounded above, and since it is increasing we would know that it converges by the MCT.

For the base case of $$ n=0$$ we have $$ a_1+2a_2\ge <!--\; \ge \; -->a_1=T_0$$ so it holds true. Now let $$ n\in <!--\; \in \; -->{\mathbb{N}}_0$$ and suppose that $$ a_1+2\underset{i=2}{\overset{2^n}{\sum <!--\; \sum \; -->}}{a}_n\ge <!--\; \ge \; -->T_n$$ we need to prove that $$ a_1+2\underset{i=2}{\overset{2^{n+1}}{\sum <!--\; \sum \; -->}}{a}_n\ge <!--\; \ge \; -->T_{n+1}$$, then notice that In this case the 3rd line follows by a similar argument to the previous induction step, but this team using the fact that for all $$ x\in <!--\; \in \; -->\{2^n+1,\dots <!--\; \dots \; -->,2^{n+1}\}$$ we have $$ a_x\ge <!--\; \ge \; -->a_{2^{n+1}}$$ because $$ \left(a_n\right)$$ is monotone decreasing.

Suppose everything in the hypothesis is true, let's prove that $$ \underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{a}_n$$ converges. Clearly we can pick an $$ r\in <!--\; \in \; -->(l,1)$$ since $$ l<<!--\; <\; -->1$$ and since $$ lim sup\left(\sqrt[n]{a_n}\right)=l$$ by using $$ \mathrm{ϵ<!--\; ϵ\; -->}=r-<!--\; -\; -->l\in <!--\; \in \; -->{\mathbb{R}}^{+}$$, then we know that there is some $$ N\in <!--\; \in \; -->{\mathbb{N}}_1$$ such that for all $$ n\ge <!--\; \ge \; -->N$$ we have $\[ |sup\left(\{\sqrt[k]{a_k}:k\ge <!--\; \ge \; -->n\}\right)-<!--\; -\; -->l|<<!--\; <\; -->\mathrm{ϵ<!--\; ϵ\; -->} \]$ since $$ \sqrt[n]{a_n}\in <!--\; \in \; -->\{\sqrt[k]{a_k}:k\ge <!--\; \ge \; -->n\}$$ then we must have $$ \sqrt[n]{a_n}\le <!--\; \le \; -->sup\left(\{\sqrt[k]{a_k}:k\ge <!--\; \ge \; -->n\}\right)$$ as it's an upper bound, thus $\[ |\sqrt[n]{a_n}-<!--\; -\; -->l|<<!--\; <\; -->|sup\left(\{\sqrt[k]{a_k}:k\ge <!--\; \ge \; -->n\}\right)-<!--\; -\; -->l|<<!--\; <\; -->\mathrm{ϵ<!--\; ϵ\; -->} \]$ so that $$ \sqrt[n]{a_n}<<!--\; <\; -->l+\mathrm{ϵ<!--\; ϵ\; -->}=l+(r-<!--\; -\; -->l)=r$$, in other words $$ a_n<<!--\; <\; -->r^n$$ for all $$ n\ge <!--\; \ge \; -->N$$, then we construct the sequence by

• $$ b_n:=a_n$$ for $$ n\in <!--\; \in \; -->[1,\dots <!--\; \dots \; -->N-<!--\; -\; -->1]$$
• $$ b_n:=r^n$$ if $$ n\ge <!--\; \ge \; -->N$$

thus by construction we have $$ \left|a_k\right|=a_k\le <!--\; \le \; -->b_k$$ for all $$ k\in <!--\; \in \; -->{\mathbb{N}}_1$$ moreover we can confirm that $$ \left(b_n\right)$$ is summable directly as $\[ \underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{b}_n=\underset{n=1}{\overset{N-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}{b}_n+\underset{n=N}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{r}^n=\underset{n=1}{\overset{N-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}{b}_n+\frac{r^N}{1-<!--\; -\; -->r} \]$ the final equality comes from a geometric series with an offset since the right hand side is the sum of two finite things then $$ \left(b_n\right)$$ is summable and thus by the comparison test we conclude that $$ \underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{a}_n$$ converges.

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Now suppose that $$ $$