How come this series is convergent?

Question

How does the below series converges

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

I used ratio test and got 1 (which means divergent). But i was told the series is convergent, i don't understand.

Edit: Thank you all. I understand now. :)

Answer 1

No the 1 you got means it's inconclusive (no conclusion possible with this test) and you need to use another method

Answer 2

The ratio test fails if you get the value as 1, for a counter example $\sum_{n=1}^\infty\frac{1}{n}$ and $\sum_{n=1}^\infty\frac{1}{n^2}$ has 1 as limit of ratios where the former series diverges and the latter one converges. To know why $\sum_{n=1}^\infty\frac{(-1)^n}{n}$ converges you need to know a theorem on "alternating series" by Leibnitz. This theorem is stated as follows:

Suppose
(a) $\left\lvert c_1 \right\rvert$ $\geqslant$ $\left\lvert c_2 \right\rvert$ $\geqslant$ $\left\lvert c_3 \right\rvert$ . . . ;
(b) $c_{2m-1}$ $\leqslant$ 0 , $c_{2m}$ $\geqslant$ 0 (m=1 , 2 , 3 ,. . .);
(c) $\lim_{n\to \infty}$ $c_{n}$ = 0

Then $\sum_{n=1}^\infty c_n$ converges.

Here your $c_n$ = $\frac{(-1)^n}{n}$, which clearly satisfies all the above mentioned condition. Thats the reason why $\sum_{n=1}^\infty\frac{(-1)^n}{n}$ converges.

Answer 3

As a hint: $\frac1n\to0$ and $$(-1)\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\\=\frac11-\frac 1{2}+\frac 13-\frac 14+\frac 15-\frac 16+\cdots\\=\left(\frac11-\frac 1{2}\right)+\left(\frac 13-\frac 14\right)+\left(\frac 15-\frac 16\right)+\cdots\\=\frac 12 +\frac 1{12}+\frac 1{30}+\cdots\\<\left(\frac11-\frac 12\right)+\left(\frac12-\frac 13\right)+\left(\frac13-\frac 14\right)+\cdots=1.$$