Are series test flawed?

[TMarvel]

I solved a infinite power series which is convergent, but Wolfram Alpha is concluding its divergent through many tests (geometric test, limit test, etc..).
So, are those tests not for all power series? Or are they flawed?
And this series is no exception, I can give infinite amount of similar series if you want!
Anyways, here the series.
[math]S=\sum_{k=0}^{∞}{\frac{{(-4)}^{k}}{{3}^{k+1}}} \\ S= \frac{1}{3}-\frac{{4}}{{3}^{2}}+\frac{{4}^{2}}{{3}^{3}}-\frac{{4}^{3}}{{3}^{4}}+... [eq(1)]\\ 3S= 1-\frac{{4}}{{3}^{}}+\frac{{4}^{2}}{{3}^{2}}-\frac{{4}^{3}}{{3}^{3}}+...\\ 1-3S=\frac{{4}}{{3}^{}}-\frac{{4}^{2}}{{3}^{2}}+\frac{{4}^{3}}{{3}^{3}}-...\\ \frac{1-3S}{4}=\frac{{1}}{{3}^{}}-\frac{{4}^{}}{{3}^{2}}+\frac{{4}^{2}}{{3}^{3}}-...[eq(2)]\\ \\ eq(1)-eq(2)\\ \Rightarrow S-\frac{1-3S}{4}=0\\ 4S-1+3S =0\\ 7S=1\\ S=1/7\\ ∴ the\ series\ is\ convergent[/math]Here's a screenshot of Wolfram Aplha

Note: An amateur here so pls try to keep it simple

 

[blamocur]

It's pretty neat, but unfortunately you first have to prove that the sum converges. Consider a simpler example: [imath]S = 1 + 2 + 2^2 + 2^3 + ...[/imath], i.e. [imath]S = \sum_{k=0}^\infty 2^k[/imath]. Using your trick we can show that [imath]S = 1 + 2S[/imath] and thus [imath]S = -1[/imath]

 

[Otis]

Hi TM. By manipulating infinite sums like that, we can show also that the set of Natural numbers sums to -1/12.

? What does that say about the method.

[imath]\;[/imath]

 

[Dr.Peterson]

I solved a infinite power series which is convergent, but Wolfram Alpha is concluding its divergent through many tests (geometric test, limit test, etc..).
So, are those tests not for all power series? Or are they flawed?
And this series is no exception, I can give infinite amount of similar series if you want!
Anyways, here the series.
[math]S=\sum_{k=0}^{∞}{\frac{{(-4)}^{k}}{{3}^{k+1}}} \\ S= \frac{1}{3}-\frac{{4}}{{3}^{2}}+\frac{{4}^{2}}{{3}^{3}}-\frac{{4}^{3}}{{3}^{4}}+... [eq(1)]\\ 3S= 1-\frac{{4}}{{3}^{}}+\frac{{4}^{2}}{{3}^{2}}-\frac{{4}^{3}}{{3}^{3}}+...\\ 1-3S=\frac{{4}}{{3}^{}}-\frac{{4}^{2}}{{3}^{2}}+\frac{{4}^{3}}{{3}^{3}}-...\\ \frac{1-3S}{4}=\frac{{1}}{{3}^{}}-\frac{{4}^{}}{{3}^{2}}+\frac{{4}^{2}}{{3}^{3}}-...[eq(2)]\\ \\ eq(1)-eq(2)\\ \Rightarrow S-\frac{1-3S}{4}=0\\ 4S-1+3S =0\\ 7S=1\\ S=1/7\\ ∴ the\ series\ is\ convergent[/math]Here's a screenshot of Wolfram Aplha

Note: An amateur here so pls try to keep it simple

What you've really shown is just that if the series converges (so that it is valid to manipulate it at all), then its sum is 1/7.

That's the reason for convergence tests! In this case, the tests fail, so you can't do any of your work. You don't do this sort of work first, and claim that it proves convergence.

Do you honestly believe that the sequence of partial sums 1/3, -1/9, 13/27, -25/81, 181/243, -481/729, 2653/2187, ... converges to 1/7??? Have you tried calculating them? That's the first thing an amateur should do.

 

[lookagain]

TMarvel, the absolute value of any term in this series is greater than its predecessor here. For an alternating series that is geometric to be convergent, it is necessary for the absolute values of the succeeding terms to be smaller than their predecessors.
That is not a statement of sufficiency.

This series fails the necessary condition. Or, imagine that you multiply both sides
of your equation (1) to get:

\(\displaystyle 4S \ = \ \dfrac{4}{3} \ - \ \dfrac{4^2}{3^2} \ + \ \dfrac{4^3}{3^3} \ - \ ...\)

Here, the ratio, r, is -4/3. Its absolute value needs to be less than 1 for the series
to be convergent.

 

[TMarvel]

It's pretty neat, but unfortunately you first have to prove that the sum converges. Consider a simpler example: [imath]S = 1 + 2 + 2^2 + 2^3 + ...[/imath], i.e. [imath]S = \sum_{k=0}^\infty 2^k[/imath]. Using your trick we can show that [imath]S = 1 + 2S[/imath] and thus [imath]S = -1[/imath]

Yeah, you're right! This method is kinda misleading!

 

[TMarvel]

Hi TM. By manipulating infinite sums like that, we can show also that the set of Natural numbers sums to -1/12.

? What does that say about the method.

[imath]\;[/imath]

Ya, Sir Ramanujan's summation. I wonder if these types of sum even have a meaning!?

 

[TMarvel]

What you've really shown is just that if the series converges (so that it is valid to manipulate it at all), then its sum is 1/7.

That's the reason for convergence tests! In this case, the tests fail, so you can't do any of your work. You don't do this sort of work first, and claim that it proves convergence.

Do you honestly believe that the sequence of partial sums 1/3, -1/9, 13/27, -25/81, 181/243, -481/729, 2653/2187, ... converges to 1/7??? Have you tried calculating them? That's the first thing an amateur should do.

Yeah sorry,
Actually I was experimenting with binomial theorem and found that every fraction can be written as an infinite power series(except 1) and so I got excited and did this nonsense.?

Thanks for helping though, it really helped me alot.?