# Proving $\sum_{i=1}^{\infty} \frac{2^i}{i} = 0 in \mathbb{Q}_2$

• caji
In summary: This means that \log_p(-1) = 0, which implies that -\log(-1) = 0, and thus we can conclude that:\sum_{n=1}^{\infty} \frac{2^n}{n} = 0In summary, we have shown that in \mathbb{Q}_2, the series \sum_{n=1}^{\infty} \frac{2^n}{n} converges to 0.
caji

## Homework Statement

Show, that in $$\mathbb{Q}_2$$ it holds

$$\sum_{i=1}^{\infty} \frac{2^i}{i} = 0$$

Hint: Consider $$\log_2(-1)$$

## Homework Equations

Just for completeness:

$$\log(1+s) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{s^n}{n}$$

## The Attempt at a Solution

Actually it's no homework but an exercise from a book. I found already, that

$$\lim_{i \to \infty} |2^i|_2 = 0$$

which seems to be correct as I found another thread in this forum called "p-adic convergence" which says the same. But that doesn't help me, because

$$\lim_{i \to \infty} \left|\frac{1}{i} \right|_2$$

does not converge.

Does anybody know, how to use this hint given in the exercise?

caji

Ps.: Hmm, if one looks at $$\log(-1)$$, then in the upper formula, $$s=-2$$. That means $$\sum_{i=1}^{\infty} \frac{2^i}{i} = -\log(-1) \overset{?}{=} 0$$. Could that be correct?

Last edited:

Dear caji,

Thank you for your post. It seems like you're on the right track with your thinking. Let me provide a few more steps to help guide you towards a solution.

First, let's consider the p-adic logarithm function defined as:

\log_p(x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-1)^n}{n}

Note that this is similar to the logarithm function in the homework equations, but with the addition of the p-adic absolute value on the input, which is necessary since we are working in \mathbb{Q}_2. Now, we can use the definition of the p-adic absolute value to rewrite this as:

\log_p(x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{|x-1|_p^n}{n}

Next, let's consider the p-adic logarithm of -1, denoted as \log_p(-1). Using the definition above, we can rewrite this as:

\log_p(-1) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{|(-1)-1|_p^n}{n} = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{|-2|_p^n}{n}

Now, we can use the fact that the p-adic absolute value of -2 is equal to the p-adic absolute value of 2, since |x|_p = |y|_p if and only if x and y are p-adic units (i.e. they have the same p-adic valuation). Therefore, we can rewrite this as:

\log_p(-1) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{|2|_p^n}{n} = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{2^n}{n}

Finally, we can use the fact that in \mathbb{Q}_2, the p-adic absolute value of 2 is equal to 1/2, since the p-adic valuation of 2 is equal to 0. Therefore, we can rewrite

## 1. What is the meaning of $\mathbb{Q}_2$ in this equation?

The notation $\mathbb{Q}_2$ represents the field of 2-adic numbers, which is a way of extending the rational numbers to include infinite decimal expansions with respect to the prime number 2.

## 2. How can the infinite sum of terms with non-zero values equal 0?

In the 2-adic number system, values get smaller as the terms get larger, so even though each term in the sum is non-zero, their values approach 0 as the sum goes to infinity.

## 3. Is this equation true for all values of i?

Yes, as long as i is a positive integer, the equation holds true.

## 4. Can this equation be proven using traditional methods of calculus?

No, this equation can only be proven using concepts from p-adic analysis, which is a branch of number theory that deals with the properties of p-adic numbers.

## 5. What are some real-world applications of understanding this equation?

Understanding this equation can have implications in number theory, cryptography, and coding theory. It also has connections to the study of fractals and self-similar structures.

• Calculus and Beyond Homework Help
Replies
3
Views
374
• Calculus and Beyond Homework Help
Replies
3
Views
714
• Calculus and Beyond Homework Help
Replies
6
Views
518
• Calculus and Beyond Homework Help
Replies
13
Views
230
• Calculus and Beyond Homework Help
Replies
1
Views
573
• Calculus and Beyond Homework Help
Replies
8
Views
922
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
771
• Calculus and Beyond Homework Help
Replies
4
Views
666
• Calculus and Beyond Homework Help
Replies
17
Views
2K