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.

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


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