0
$\begingroup$

There was this question in our midsem question paper:

We had to find out the range of values of x (x is positive) for which the given series is convergent,

Given series was $\sum_{n = 1}^{\infty} (a + b + c)$ where a, b and c are some functions in terms of x and n.

Now, b was something like this: $\cfrac{1}{(3n-1)^{(x+1)}}$.

Now this series is convergent for all positive real values of x.

Then, I mentioned that:

  1. For the series (a+b+c) to be convergent and b proved to be convergent, series (a+c) must also be convergent (since, sum of a convergent and divergent series cannot be convergent).

  2. Then, I applied power series formula for radius of convergence and obtained the answer.

Now, my answer and the correct answer match, but this question which is worth 18 credit points, my instructor has given me zero... What I would like to know is, whether there I any mistake in my method or not.

Thanks a lot

Vishwesh

PS: Sorry for not mentioning the problem earlier, I had to search the question paper. Here is the question:

$\sum_{n=1}^{\infty}[\cfrac{x^{3n-2}}{(3n-2)!} + \cfrac{1}{(3n-1)^{x+1}} + x^{3n}]$

Here is my solution (sorry but I will have to upload it as an image):

enter image description here enter image description here enter image description here

$\endgroup$
10
  • $\begingroup$ Was it already shown in your course that the sum of a convergent and divergent series makes a divergent series? You may have been required to justify this step. Other than that, your overall method works and I don't see any issues. $\endgroup$
    – Peter Huxford
    Oct 28, 2014 at 7:02
  • $\begingroup$ Thanks a lot for your comment. Yes, in our textbook (Thomas Calculus), there is a statement and some examples on the same : 1. Sum of two convergent series is always convergent. 2. Sum of two divergent series may be divergent. 3. Sum of a convergent and divergent series is always divergent. I just used these... $\endgroup$
    – Vishwesh
    Oct 28, 2014 at 7:02
  • $\begingroup$ You are asking expertise regarding an answer deemed faulty by your instructor, without showing it to us. I would be reluctant to formulate an opinion until you actually show the exam paper you handed. $\endgroup$
    – Did
    Oct 28, 2014 at 7:07
  • $\begingroup$ Thanks for your comment... Basically, our instructor has mentioned in our class that unless and until your answer is mathematically and logically incorrect, you can use any method you want... I am not saying that I want more credits (though I do but that is not my primary objective), I mainly want to know whether I can use this approach in future or not... (not in my exams, but as an independent question) $\endgroup$
    – Vishwesh
    Oct 28, 2014 at 7:09
  • 1
    $\begingroup$ Not sure you got the serious problem I indicated with what you posted, so let me reiterate (but only once): you show that if the limit of the ratios is $\lt1$ then $|x|\lt1$. This can prove nothing about the exact value of the radius of convergence. For example, it is equally true that if the limit of the ratios is $\lt1$ then $|x|\lt2$, which does not make the radius of convergence equal to $2$, does it? Basically, the trouble is that you show $P\implies Q$ when you would need something similar to $Q\implies P$. (And, once again, your proof says nothing about the boundary case.). $\endgroup$
    – Did
    Nov 2, 2014 at 12:15

1 Answer 1

Reset to default
0
$\begingroup$

I suspect that the devilish detail is hidden in the fact that after getting rid of $b$ you still have two summands $a$ and $c$. If you computed the radius of convergence $R_a$ for $a$ and the radius of convergence $R_c$ for $c$, then it follows that $a_n+c_n$ converges within radius $x<\min\{R_a,R_b\}$, and we have divergence for $\min\{R_a,R_c\}<x<\max\{R_a,R_c\}$. However, nothing can be said so straightfirwardly about $\max\{R_a,R_c\}<x$ (not to mention the nedd of special investicgation at $x=R_a$ and $x=R_c$). Since your argument "1" is correct and if you sufficiently reasoned that $b$ is convergent, I suppose that your answer might be worth a strictly positive number of credits. However, depending on the specific problem, this may be only a rather small part of the complete solution. (As mentioned in the comments, it is not really possible to give any conclusive statement without knowing the exact problem statement, your solution, and course requirements)

$\endgroup$
1
  • $\begingroup$ Thanks a lot @Hagen. But, I calculated radius of convergence for a+c as a whole. $\endgroup$
    – Vishwesh
    Oct 28, 2014 at 7:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .