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$$\sum_{n=1}^\infty (-1)^n\frac{n}{10^n}$$

I am trying to show that $|a_{n+1}|$ is less than or equal to $|a_n|$

So I have $\frac{n+1}{10^{n+1}}$ less than or equal to $\frac{n}{10^n}$.

And then that $\lim_{n\to\infty} a_n = 0$

Both are supposedly true and makes the series convergent but when I do it I get not true for both conditions and then divergent for the series, which is wrong.

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What do you mean "when I do it?" The series converges for precisely the reasons you said. – Adam Saltz Jul 27 '11 at 1:30
I've put your equations into LaTeX. – Zev Chonoles Jul 27 '11 at 1:31

I know there are already 4 answers, but I wanted to talk about intuition rather than any sort of exact approach here. So first, I wanted to mention that I think to those well-versed in limits, it seems readily apparent that $\lim \frac{n}{10^n} = 0$. Why? Some will mention that "exponentials grow faster than polynomials." This is true. Some say that "polynomials grow faster than logs, and this is equivalent." This is also true. But these don't build understanding.

One way to think of this would be to just start counting. 0, 1, 2, 3, 4... on one hand. 1, 10, 100, 1000... on the other hand. Ok, so it feels like $10^n$ grows faster from this sort of spotty assessment. How can we assure ourselves that this is true? A standard idea is to look at the differences of the differences. So for example, when 0 goes to 1, the difference is 1. When 1 goes to 2, the difference is 1. In fact, it's clear that for every increase by 1, the value of the function increases by 1. The difference is always 1. So the change in the difference is zero. (yes, it's a derivative of $f(x) = x$, and the second derivative too). What about the other? Well, 1 to 10 is 9. 10 to 100 is 90, which is 81 more than the previous increase. 100 to 1000 is 900, which is 810 more than the previous. And this still increases too (and if you look at more depths of differences, than those increases always increases too - that's why exponentials get so incredibly big).

But there's a way that I like better. When I think of dividing by 10, I think of "moving the decimal over 1." So each time n increases, I move the decimal over. In order to keep up, the top would have to increase quickly, but only increasing by 1 each time is too slow.

Secondly, I encourage you to always write several terms in the series out to get a feel for the numbers. I think the problem is that from 0 to 1, the ratio increases. But from then on, it decreases, and it decreases very quickly. Anyhow, good work and good luck. I hope this is useful.

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Hint: $\frac{n+1}{10^{n+1}} \le \frac{n}{10^n}$ is equivalent to $\frac{n+1}{n} \le \frac{10^{n+1}}{10^n}$

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You could just look at $|a_n|$ and apply the ratio test. Absolute convergence implies conditional convergence.

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For any $n\geq 1$, it is true that $$1<9n.$$ Adding $n$ to both sides, we get $$n+1<10n.$$ Dividing by $10^{n+1}$ on both sides, we get

$$\frac{n+1}{10^{n+1}}<\frac{n}{10^n}.$$ Thus, by the alternating series test, the series $$\sum_{n=1}^\infty (-1)^n\frac{n}{10^n}$$ converges.

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After an initial hump, the sequence $n\mapsto {n/10^n}$ decreases to zero. Apply the alternating series test.

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