0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.