# Is this power series inclusive of the endpoints?

Say I have this series:

$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n5^n}x^n$$

So if I use the ratio test:

$$|\frac{a_{n+1}}{a_n}| = \sum_{n=1}^{\infty} \frac{(-1)^nx^{n+1}}{(n+1)5^{n+1}} * \frac{n*5^n}{(-1)^{n-1} x^n} = |\frac{xn}{(n+1)5}| = |\frac{x}{5}|$$

So I used the ratio test to determine that the interval of convergence is $$-5 < x < 5$$ before checking the endpoints.

But if I plug in x = -5, I get a werid series:

$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n5^n} *-5^n$$

I can't use the alternating series test to show convergence since b = $$\frac{-5^n}{n5^n}$$isn't positive right?. What can I do?

When x = 5, this converges by alternating series test and it's similar to a harmonic series.

• When I plug in $x=-5$, I get a non-weird series $$\sum_{n=1}^\infty\frac{-1}n.$$ – Lord Shark the Unknown Apr 10 at 4:37
• But the answer apparently using a calculator says it includes both -5 adn 5 in the interval of convergence. – Jwan622 Apr 10 at 14:22
• What on earth does "the answer apparently using a calculator" mean? – Lord Shark the Unknown Apr 10 at 14:52
• – Jwan622 Apr 10 at 15:15
• Well, whatever that is, and whatever that says, your series diverges when $x=-5$. – Lord Shark the Unknown Apr 10 at 15:54

You have a small notation error: "$$-5^n$$". This is always negative, but powers of $$x$$ when $$x = -5$$ should alternate sign, depending on whether that power is even or odd.
Note that \begin{align*} \left. \frac{(-1)^{n-1}}{n 5^n} x^n \right|_{x=-5} &= \frac{(-1)^{n-1}}{n 5^n} (-5)^n \\ &= \frac{(-1)^{n-1}}{n 5^n} (-1)^n 5^n \\ &= \frac{(-1)^{2n-1}}{n} \\ &= \frac{-1}{n} \end{align*} It should be fairly clear that minus the harmonic series diverges.