# Does $\lim\limits_{n \rightarrow \infty} \sum_{i=1}^n 4 \cdot 5^i - 5^{n+1}=-1$?

I first ran into the formula:

$$S = \lim_{n \rightarrow \infty} \sum_{i=1}^n 4 \cdot 5^i - 5^{n+1} = -1$$

in the context of $$p-$$adics, where we define a new norm, where for $$x = 5^a \cdot b$$, $$5 \not \lvert b$$, the norm is defined as $$||x|| \equiv 5^{-a}$$. That is, $$a$$ is the largest power of $$5$$ in the prime factorization of $$x$$.

Under this definition, we can see that $$\lim_{n\rightarrow \infty}||5^{n+1}|| = 0$$, and $$\lim_{n\rightarrow \infty} \sum_{i=1}^n 4 \cdot 5^n = \frac{4}{1-5} = -1$$. The GP formula is legal since $$||5|| = 5^{-1} < 1$$.

Hence, the total limit is $$-1 - 0 = -1$$.

### Using the same formula under the usual norm

However, let's now decide to use the $$\epsilon-\delta$$ definition of convergence of a series, under the usual norm $$|x| = abs(x)$$ (the absolute value). Now, we claim that the limit of the series $$S$$ is $$-1$$. That is, we need to show that

$$A_n \equiv \sum_{i=1}^n 4 \cdot 5^i - 5^{n+1} \\ \forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n \geq N, |A_n - (-1)|< \epsilon$$

Let's evaluate $$|A_n - (-1)|$$:

\begin{align*} |A_n - (-1)| &= |A_n + 1| = |4 + 4\cdot5 + 4 \cdot 5^2 + \dots + 4 \cdot 5^n - 5^{n+1} + 1| \\ &= |5 + 4 \cdot 5 + 4 \cdot5^2 + \cdots 4\cdot 5^n - 5^{n+1}| \\ &= |5\cdot 5+ 4\cdot 5^2 + \cdots + 4\cdot 5^n - 5^{n+1}| \\ &= |5^2 + 4\cdot 5^2 + \cdots + 4\cdot 5^n - 5^{n+1}| \\ &= |5\cdot 5^2 + \cdots + 4\cdot 5^n - 5^{n+1}| \\ &= \cdots \\ &= |5\cdot 5^n - 5^{n+1}| \\ &= |0| = 0 \end{align*}

That seems to imply that the series:

$$S = \lim_{n \rightarrow \infty} \sum_{i=1}^n 4 \cdot 5^i - 5^{n+1} = -1$$

under the usual norm! But this makes no sense, the terms keep getting bigger, there is no reason this should converge? What am I missing?

If it does indeed converge, then why do we need the $$p-$$adic norm to explain this situation?

• $\sum_{i=1}^n( 4 \cdot 5^n - 5^{n+1}) = n ( 4 \cdot 5^n - 5^{n+1})$ since you are summing $n$ identical summands. Do you mean $\sum_{i=1}^n( 4 \cdot 5^i - 5^{i+1})$? Jan 15, 2020 at 21:29
• $4\cdot5^i-5^{i+1}=-5^i$; do you mean $\left(\sum4\cdot5^i\right)-5^{n+1}$ or $\sum\left(4\cdot5^i-5^{i+1}\right)$ Jan 15, 2020 at 21:29
• @J.W.Tanner: I fixed the bug. I did indeed mean $(\sum 4 \cdot 5^i) - 5^{n+1}$. Jan 15, 2020 at 21:41

I think you're conflating a couple of things here. Let $$S_n$$ be the sequence of partial sums of the geometric series, $$S_n=\sum_{i=0}^n4\cdot 5^i$$. Then it is true for all $$n$$ that $$S_n=5^{n+1}-1$$ by the usual formula, so that if we look at the 'associated' sequence $$\mathfrak{S}_n=S_n-5^{n+1}$$, then $$\mathfrak{S}_n$$ is $$-1$$ for all $$n$$, so $$\lim\limits_{n\to\infty}\mathfrak{S}_n = -1$$ because it's the limit of a constant sequence.
But what makes the discussion interesting in the context of the $$5$$-adics isn't that this limit of $$\mathfrak{S}_n$$ converges; rather, by the explicit formula we have $$|S_n-(-1)|_5$$ = $$|5^{n+1}|_5$$ $$=5^{-(n+1)}$$. And since the norm of the difference between $$S_n$$ and $$-1$$ converges to zero as $$n\to\infty$$, the limit of the $$S_n$$ — that is, the limit of the sum itself — exists in the $$5$$-adic topology, and there $$\lim\limits_{n\to\infty}S_n=-1$$, so we can assign a meaningful value to the sum $$\sum_{i=0}^\infty 4\cdot 5^i$$.
• Thanks a lot! I now understand why the $p-$adic result is interesting! Jan 16, 2020 at 6:49