7
$\begingroup$

Consider the series: $$\sum_{n=1}^\infty\frac{(x-3)^n}{n}$$ Doing hand-calculations, I applied the Ratio Test and found that it converges if $|x-3|<1$, making the radius of convergence $R=1$.

Next, I used hand-calculations to test the endpoints. Substituting $x=2$, the series becomes $$\sum_{n=1}^\infty\frac{(2-3)^n}{n}=\sum_{n=1}^\infty\frac{(-1)^n}{n},$$ which is an alternating series that converges. Substituting $x=4$, the series becomes $$\sum_{n=1}^\infty\frac{(4-3)^n}{n}=\sum_{n=1}^\infty\frac{1}{n},$$ which is a harmonic series that diverges. Therefore, the interval of convergence is $[2,4)$.

To check my answer, I used:

SumConvergence[(x - 3)^n/n, n]

Which returns:

Abs[-3 + x] <= 1 && x != 4

Which agrees with all of my hand-calculations. I got pretty excited about this, as this would really help my students check their work. I was thinking it will not only give the radius of convergence, but check the endpoints.

But then I examined the series $$\sum\frac{x^n}{\sqrt{n}}$$

My hand-calculations returned that the series converges if $|x|<1$. I then checked the endpoints of the interval and was able to determine convergence at $x=-1$, but divergence at $x=1$, making the interval of convergence $[-1,1)$. I then tried to check my answer with SumConvergence. The command

SumConvergence[x^n/Sqrt[n], n]

returned:

Abs[x] < 1

But then I tried

SumConvergence[(-1)^n/Sqrt[n], n]

which returned:

True

And I tried

SumConvergence[(1)^n/Sqrt[n], n]

which returned:

False

I am using Mathematica 11.1.0.0. So, my question, how come

SumConvergence[x^n/Sqrt[n], n]

did not return

Abs[x] < 1 && x != 1

especially since SumConvergence[(-1)^n/Sqrt[n], n] returned "True" and SumConvergence[(1)^n/Sqrt[n], n] returned "False."

Additional example to show it is not just a square root problem:

SumConvergence[((-1)^(n - 1) x^n)/n^3, n]

Returns:

Abs[x] < 1

It should be:

Abs[x] <= 1

Because SumConvergence[((-1)^(n - 1) (-1)^n)/n^3, n] and SumConvergence[((-1)^(n - 1) (1)^n)/n^3, n] both return "True."

Improve this question
$\endgroup$
4
  • $\begingroup$ I think there is a mistake in the expected result, shouldn't it be Abs[x] < 1 && x ==-1? $\endgroup$ Commented Mar 25, 2017 at 3:06
  • $\begingroup$ @AnjanKumar I was emulating this return above: Abs[-3 + x] <= 1 && x != 4 $\endgroup$ Commented Mar 25, 2017 at 3:21
  • $\begingroup$ When I run SumConvergence[x^n/n, n] it returns correct result, whereas, SumConvergence[x^n/Sqrt[n^2], n] returns Abs[x] < 1. May be it has got something to do with Sqrt. This looks like a bug. $\endgroup$ Commented Mar 25, 2017 at 3:38
  • $\begingroup$ @AnjanKumar See my example in my update. Probably not just a square root problem. $\endgroup$ Commented Mar 25, 2017 at 3:50

1 Answer 1

Reset to default
3
$\begingroup$

I'm not sure what the question is -- whether it's, "Does SumConvergence return only generically true results like some other solvers?", or "Is this a bug?" -- but in V11.2, we get correct and complete results:

SumConvergence[(x - 3)^n/n, n] (* Abs[-3 + x] <= 1 && x != 4 *) SumConvergence[x^n/Sqrt[n], n] (* Abs[x] < 1 || x == -1 *) SumConvergence[((-1)^(n - 1) x^n)/n^3, n] (* Abs[x] < 1 || x == 1 || x == -1 *)
Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.