I have all the questions correct on my hw except for one: find where $\sum_1^{\infty} (x+4)^{n}$ converges conditionally.

Radius of Convergence

I got 1 for this, by using the root test and finding the interval.

Interval of Convergence

Using the root test, I got:



So the interval of convergence would be (-5,-3) (we have to write it in interval form)

Where does the series converge absolutely? (also explain this to me)

So for this one, I tested the endpoints.

At -5: $\sum_1^{\infty} |(-5+4)^{n}|$ = $|-1|^{n}$ = $|1|^{n}$ diverges

At -3: $\sum_1^{\infty} |(-3+4)^{n}|$ = $|1|^{n}$ diverges

So I did not know how to write the interval for this one, so I wrote (-5,-3) and my hw said it was correct but if someone can explain this logic to me I would appreciate it.

Where does the series converge conditionally? (also explain this to me)

So for this one I tested the endpoints normally without absolute value bars.

At -5: $\sum_1^{\infty} (-5+4)^{n}$ = $(-1)^{n}$ diverges

At -3: $\sum_1^{\infty} (-3+4)^{n}$ = $1^{n}$ diverges

So I do not know how to write this interval.

Help would be appreciated!!! Thank you!

  • $\begingroup$ Your series converges absolutely for $-5\lt x\lt -3$, and diverges for $x\lt -5$ and for $x\gt -3$. It also diverges at $x=-5$ and $x=-3$, so it converges conditionally nowhere. $\endgroup$ – André Nicolas Apr 23 '15 at 6:02
  • $\begingroup$ @AndréNicolas Does it not converge at $x=-4$? $\endgroup$ – MathMajor Apr 23 '15 at 6:06
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    $\begingroup$ @Elsa It may converge absolutely on one interval and diverge elsewhere. $\endgroup$ – MathMajor Apr 23 '15 at 6:09
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    $\begingroup$ Absolute convergence means the sum of the absolute values converges. It implies convergence. The Ratio Test and Root Test tell you nothing about what happens at the endpoints, You could have absolute convergence at the endpoints, or divergence, or conditional convergence at one or both endpoints. $\endgroup$ – André Nicolas Apr 23 '15 at 6:28
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    $\begingroup$ What you did, except that in addition we have to test for absolute convergence at the endpoints. In this case, we have that the sum of the absolute values diverges at the endpoints, so the interval of absolute convergence is $(-5,-3)$. $\endgroup$ – André Nicolas Apr 23 '15 at 6:40

For the first part, you have established that $-5$ and $-3$ are not part of the radius of convergence, therefore the interval is $(-5,\,3)$, and not $[-5,\,3]$ (which would include $-5$ and $-3$).

Same logic for the second part. You've shown that the series converges for $-5 < x < -3$ but diverges for $-5$ and $-3$, therefore, it is the open interval $(-5,\,-3)$.

Recall that

$$(a,\,b) = \{x \mid a<x<b\}$$ whereas $$[a,\,b] = \{x \mid a \le x \le b \}.$$

  • $\begingroup$ i already tried (-5,-3) for the conditionally converging one but it said it was wrong. $\endgroup$ – Elsa Apr 23 '15 at 6:01
  • $\begingroup$ Ah, there must be a flaw in your test. I admit that I did not look that carefully. $\endgroup$ – MathMajor Apr 23 '15 at 6:07
  • $\begingroup$ ok can u explain also how my series converges absolutely if both of the endpoints diverge? like if both endpoints converged would it be [-5,-3] for converging absolutely? i dont get how if it diverged the interval would still be valid. also for converging conditionally i still do not get it. $\endgroup$ – Elsa Apr 23 '15 at 6:10
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    $\begingroup$ Yes, that is correct. So if the endpoints diverge than we exclude them, i.e. $(-5,\,-3)$. If the endpoints converge then we would have $[-5,\,-3]$. It may be possible that only the left endpoint converges. In that case, we would have $[-5,\,-3)$. $\endgroup$ – MathMajor Apr 23 '15 at 6:12
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    $\begingroup$ I'm not sure, are you entering it in correctly? WolframAlpha gives convergence on $|x+4|<1$ ... $\endgroup$ – MathMajor Apr 23 '15 at 6:21

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