Jtdylktuy

I have this series: $sum_{i=1}^n frac{(-2a)^n}{n^2}$ and the question is for what values of $a$ does the series converge. I'm supposed to write the interval $a$ is in. I first tried the ratio test:

$$ frac{(-2a)^{n+1}}{(n+1)^2} cdot frac{(n)^2}{(-2a)^n} = frac{-2a cdot n^2}{(n+1)^2} = frac{-2a cdot n^2}{n^2+2n+1}.$$

I then divided everything with $n^2$:

$$ frac{-2a cdot n^2}{n^2+2n+1} = frac{-2a}{1+2/n+1/n^2},$$

and when I took the limit as $n to infty$, I ended up with the inequality:

$$ -2a < 1. $$

I divided with $-2$ on both side, and got that $a$ is $-1/2$, but I'm supposed to make an interval, what did I do wrong?

You want to take absolute values in the ratio test. The result should be that it converges if $|-2a| < 1$, i.e.
$-1/2 < a < 1/2$, and diverges if $|-2a| > 1$. But you also need to look at the case $|-2a|=1$, where the ratio test is inconclusive.

We need that

$$frac{(-2a)^n}{n^2} to 0 implies |2a|le 1$$

then

• for $-1<2a<0$ the series converges by comparison test with $sum frac1{n^2}$




• for $0<2a<1$ the series converges by alternating series test