$\Theta \rho ϵ\eta \Pi \alpha \tau \pi$

We start by proving that $1⟹2$ so assume that series converges. Let $S_n:={\sum }_{k=1}^n{a}_k$ be the sequence of partial sums, therefore we know that for any $ϵ$ there is an $N\in {\mathbb{N}}_0$ such that for all $n\ge N,|S_n-L|<ϵ$, note that $$\begin{array}{cccc}& L-S_n& amp;=(\underset{m\to \infty }{\lim }{S}_m)-S_n& \text{<span class="tml-eqn"></span>}\\ & & amp;=\underset{m\to \infty }{\lim }(S_m-S_n)& \text{<span class="tml-eqn"></span>}\\ & & amp;=\underset{m\to \infty }{\lim }\sum _{k=n+1}^m{a}_k& \text{<span class="tml-eqn"></span>}\\ & & amp;=\sum _{k=n+1}^{\infty }{a}_k& \text{<span class="tml-eqn"></span>}\end{array}$$ then it follows that $$\left|\sum _{k=n+1}^{\infty }{a}_k\right|=|S_n-L|<ϵ$$ as needed.

---

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

Now let $a,b\ge N$ and we'll prove that $\left|{\sum }_{k=a+1}^b{a}_k\right|<ϵ$, to do this note the following $$\begin{array}{cccc}& \left|\sum _{k=a+1}^b{a}_k\right|& amp;=|\sum _{k=a+1}^b{a}_k+\sum _{k=b+1}^{\infty }{a}_k-\sum _{k=b+1}^{\infty }{a}_k|& \text{<span class="tml-eqn"></span>}\\ & & amp;=|\sum _{k=a+1}^{\infty }{a}_k-\sum _{k=b+1}^{\infty }{a}_k|& \text{<span class="tml-eqn"></span>}\\ & & amp;\le \left|\sum _{k=a+1}^{\infty }{a}_k\right|+\left|\sum _{k=b+1}^{\infty }{a}_k\right|& \text{<span class="tml-eqn"></span>}\\ & & amp;<\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ& \text{<span class="tml-eqn"></span>}\end{array}$$ as needed.

---

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 $$\left|\sum _{k=n+1}^m\right|=|S_m-S_n|$$ 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 ${\sum }_{n=1}^{\infty }{a}_n$ 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 $ϵ\in {ℝ}^{+}$, also since $\left(b_n\right)$ is summable, by $3$ of the cauchy criterion we obtain some $N\in {\mathbb{N}}_1$ such that for any $n,m\ge N$ we have $\left|{\sum }_{k=n+1}^m{b}_k\right|<ϵ$, but recall the inequalities we've assumed between $\left(a_n\right),\left(b_n\right)$, which shows us that $$\left|\sum _{k=n+1}^m{a}_k\right|\le \sum _{n=k+1}^m\left|a_k\right|\le \sum _{k=n+1}^m{b}_k<ϵ$$ The first $\le$ follows from the generalized triangle inequality, the second follows from our assumed inequality. Thus $\left(a_n\right)$ is summable.

---

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 $S_n={\sum }_{k=1}^n{a}_k$ and $T_n={\sum }_{k=0}^n{2}^k{a}_{2^k}$.

Assume that ${\sum }_{n=0}^{\infty }{2}^n{a}_{2^n}$ 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 $i\in {\mathbb{N}}_0,T_i\le M$, 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 $$S_{2^n-1}\le t_{n-1}\le M$$

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

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

$$\begin{array}{cccc}& S_{2^{n+1}-1}& amp;=S_{2^n-1}+\sum _{i=2^n}^{2^{n+1}-1}{a}_i& \text{<span class="tml-eqn"></span>}\\ & & amp;\le T_{n-1}+\sum _{i=2^n}^{2^{n+1}-1}{a}_i& \text{<span class="tml-eqn"></span>}\\ & & amp;\le T_{n-1}+2^n{a}_{2^n}& \text{<span class="tml-eqn"></span>}\\ & & amp;=T_{n-1}+(T_n-T_{n-1})& \text{<span class="tml-eqn"></span>}\\ & & amp;=T_n\le M& \text{<span class="tml-eqn"></span>}\end{array}$$ 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 $x\in \{2_n,\dots ,2^{n+1}-1\}$ that $a_{2^n}\ge a_x$ 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 ${\sum }_{n=1}^{\infty }{a}_n<\infty$

Now let's assume that ${\sum }_{n=0}^{\infty }{2}^n{a}_{2^n}$ 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 $$a_1+2\sum _{i=2}^{2^n}{a}_n\ge T_n$$ Since $\left(S_n\right)$ is bounded above by some $M\in ℝ$ then by adding $a_1$ to both sides we obtain $2S_{2^n}\ge a_1+T_n$ which is to say that $2M\ge a_1+T_n$ 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 $a_1+2a_2\ge a_1=T_0$ so it holds true. Now let $n\in {\mathbb{N}}_0$ and suppose that $a_1+2{\sum }_{i=2}^{2^n}{a}_n\ge T_n$ we need to prove that $a_1+2{\sum }_{i=2}^{2^{n+1}}{a}_n\ge T_{n+1}$, then notice that $$\begin{array}{cccc}& a_1+2\sum _{i=2}^{2^{n+1}}{a}_n& amp;=a_1+2\sum _{i=2}^{2^n}{a}_n+2\sum _{i=2^n+1}^{2^{n+1}}{a}_n& \text{<span class="tml-eqn"></span>}\\ & & amp;\ge T_n+2\sum _{i=2^n+1}^{2^{n+1}}{a}_n& \text{<span class="tml-eqn"></span>}\\ & & amp;\ge T_n+2\left(2^n{a}_{2^{n+1}}\right)& \text{<span class="tml-eqn"></span>}\\ & & amp;=T_n+2^{n+1}{a}_{2^{n+1}}& \text{<span class="tml-eqn"></span>}\\ & & amp;=T_n+(T_{n+1}-T_n)& \text{<span class="tml-eqn"></span>}\\ & & amp;=T_{n+1}& \text{<span class="tml-eqn"></span>}\end{array}$$ 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 $x\in \{2^n+1,\dots ,2^{n+1}\}$ we have $a_x\ge a_{2^{n+1}}$ because $\left(a_n\right)$ is monotone decreasing.

Suppose everything in the hypothesis is true, let's prove that ${\sum }_{n=1}^{\infty }{a}_n$ converges. Clearly we can pick an $r\in (l,1)$ since $l<1$ and since $\mathrm{lim\hspace{0.17em}sup}\left(\sqrt[n]{a_n}\right)=l$ by using $ϵ=r-l\in {ℝ}^{+}$, then we know that there is some $N\in {\mathbb{N}}_1$ such that for all $n\ge N$ we have $$|\sup \left(\{\sqrt[k]{a_k}:k\ge n\}\right)-l|<ϵ$$ since $\sqrt[n]{a_n}\in \{\sqrt[k]{a_k}:k\ge n\}$ then we must have $\sqrt[n]{a_n}\le \sup \left(\{\sqrt[k]{a_k}:k\ge n\}\right)$ as it's an upper bound, thus $$|\sqrt[n]{a_n}-l|<|\sup \left(\{\sqrt[k]{a_k}:k\ge n\}\right)-l|<ϵ$$ so that $\sqrt[n]{a_n}<l+ϵ=l+(r-l)=r$, in other words $a_n<r^n$ for all $n\ge N$, then we construct the sequence by

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

thus by construction we have $\left|a_k\right|=a_k\le b_k$ for all $k\in {\mathbb{N}}_1$ moreover we can confirm that $\left(b_n\right)$ is summable directly as $$\sum _{n=1}^{\infty }{b}_n=\sum _{n=1}^{N-1}{b}_n+\sum _{n=N}^{\infty }{r}^n=\sum _{n=1}^{N-1}{b}_n+\frac{r^N}{1-r}$$ 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 ${\sum }_{n=1}^{\infty }{a}_n$ converges.

---