Question about proof with sums

Question

I am confused on how to write a formal proof for sum notations. How would I write a formal proof for this example?

Prove that $$\sum\limits_{k = 0}^\infty\frac{2}{3^k} = 3.$$ Prove that for any $\alpha \in \{0, 2\}^\mathbb{N}$ that $$0 \le \sum\limits_{k = 0}^\infty\frac{\alpha(k)}{3^k} \le 3.$$

Answer 1

It all really depends on how many theorems you have at your disposal. Since you have $\sum_{k=0}^\infty \frac{2}{3^k} = 3$ already, you know by the comparison theorem that $$\sum_{k=0}^\infty \frac{\alpha(k)}{3^k} \le \sum_{k=0}^\infty \frac{2}{3^k} = 3$$ and it is positive since all of the terms are positive.

If you can't use the comparison theorem, then you could notice that $$0 \le \sum_{k=0}^N \frac{\alpha(k)}{3^k} \le \sum_{k=0}^N \frac{2}{3^k} < 3$$

Since $$ \sum_{k=0}^N \frac{\alpha(k)}{3^k}$$ is a monotonic increasing sequence that is bounded above, it converges. At the same time we have shown that its limit is less than $$\sum_{k=0}^\infty \frac{2}{3^k} = 3.$$

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Finally to prove the convergence of the first series notice that the partial sums satisfies: $$\sum_{k=0}^N \frac{2}{3^k} = 2 \cdot \frac{1-(1/3)^{N+1}}{1-(1/3)}$$

In the limit, the term $(1/3)^N \to 0$, which means the series converges to $$\frac{2}{1-(1/3)} = 3$$

Answer 2

The series $S:=\sum_{k=0}^{\infty} \frac{2}{3^k}$ is a geometric series with first term $a=2$ and multiplier $r=1/3$, therefor $$ \lim_{n \to \infty} \sum_{k=0}^n \frac{2}{3^k} = \frac{a}{1-r} = \frac{2}{2/3} = 3. $$

Now let $T:=\sum_{k=0}^{\infty} \frac{\alpha(k)}{3^k}$. For any $k \in \mathbb N$ it's clear that $0 \leq \frac{\alpha(k)}{3^k} \leq \frac{2}{3^K}$ and so we have the following inequality involving the partial sums $T_n:=\sum_{k=0}^n \frac{\alpha(k)}{3^k}$ and $S_n:=\sum_{k=0}^n \frac{2}{3^k}$: $$ 0 \leq T_n \leq S_n $$ for all $n \in \mathbb N$. The sequence $(T_n)_{n \in \mathbb N}$ is increasing and bound from above, hence converges to some value less than $\lim_{n \to \infty} S_n = 3$. This establishes that the series $T$ itself converges to some value between $0$ and $3$, as desired.