Proving Convergence of a Series: A Guide for Mathematicians

[twoflower]

Hi, I don't know how to prove convergence of this series:

$$\sum_{n=1}^{\infty} \log \left( 1 + \frac{1}{2\sqrt{n^3}}\right)$$

I encountered this problem while solving uniform convergence in today's analysis test and I think I didn't write it correctly there..

Could you help me please?

Thank you.

 

[arildno]

Have you tried the comparison test with integrals?
From what I can see, this shows that the sum converges.

 

[twoflower]

Have you tried the comparison test with integrals?

What is it? Maybe I don't know it..I only know integral test, but the integral I'd have to compute seemed so awfully that I didn't even consider trying it.

 

[twoflower]

Well, it is the integral test I was talking about.

Let your integral be
$$\int_{1}^{\infty}ln(1+\frac{1}{2x^{\frac{3}{2}}})dx$$
Your sum will diverge or converge with this integral, agreed?

Yes, I know that (ok, I know that it exists but never used it). But I don't evaluate myself so much that I'm able to solve this integral

 

[arildno]

Try integration by parts:
You then get:
$$I=xln(1+\frac{1}{2}x^{-\frac{3}{2}})\mid_{1}^{\infty}-\int_{1}^{\infty}(\frac{x}{1+\frac{1}{2}x^{-\frac{3}{2}}})(-\frac{3}{4}x^{-\frac{5}{2}})dx$$
$$=xln(1+\frac{1}{2}x^{-\frac{3}{2}})|_{1}^{\infty}+\frac{3}{4}\int_{1}^{\infty}\frac{x}{x^{\frac{5}{2}}+\frac{1}{2}x}dx$$

Now, your critical steps will be to show that:

a) $$\infty*ln(1+\frac{1}{2}(\infty)^{-\frac{3}{2}})<\infty$$
and
b)$$\int_{1}^{\infty}\frac{x}{x^{\frac{5}{2}}+\frac{1}{2}x}dx<\infty$$

All right?

 

[twoflower]

Well I ended with slightly different integral:

$$\frac{3}{4} \int \frac{x}{x^{\frac{5}{2}} + 2x^4}\ dx$$

which I can't solve. I tried substitution:

$$\frac{3}{4} \int \frac{x}{x^{\frac{5}{2}} + 2x^4}\ dx = \frac{3}{4} \int \frac{dx}{\sqrt{x^2} + 2x^3}$$

$$t = \sqrt{x^3}$$

$$dx = \frac{2}{3}t^{-\frac{1}{3}}$$

$$... = \frac{1}{2} \int \frac{1}{\sqrt[3]{t}\left(t+2t^2\right)}$$

Which I can't solve..

 

[arildno]

Well, your integral is wrong (check it up again!), but it doesn't matter:

Do you agree that:
a) An upper bound of the integrand (the function inside the integral sign) in your case is $$\frac{1}{x^{\frac{3}{2}}}$$?
b) What does this tell you about your integral?

As you can see, there's no need to actually solve your integral, only to bound it.

 

[twoflower]

Well, your integral is wrong (check it up again!), but it doesn't matter:

Do you agree that:
a) An upper bound of the integrand (the function inside the integral sign) in your case is $$\frac{1}{x^{\frac{3}{2}}}$$?
b) What does this tell you about your integral?

As you can see, there's no need to actually solve your integral, only to bound it.

Ok, I agree with a). However, I don't know what it says me about the integral, probably that it will be finite, but I'd need some more formal way how to see that.

Anyway, how could I prove this part?

$$\infty*ln(1+\frac{1}{2}(\infty)^{-\frac{3}{2}})<\infty$$

 

[arildno]

1. Let us use my integrand, simplified to $$\frac{1}{x^{\frac{3}{2}}+\frac{1}{2}}<\frac{1}{x^{\frac{3}{2}}}$$
where the inequality should be obviously true.
But then, for any finite "b", we have:
$$\int_{1}^{b}\frac{dx}{x^{\frac{3}{2}}+\frac{1}{2}}<\int_{1}^{b}\frac{dx}{x^{\frac{3}{2}}}=2(1-\frac{1}{\sqrt{b}})$$
Thus, we have:
$$\int_{1}^{\infty}\frac{dx}{x^{\frac{3}{2}}+\frac{1}{2}}=\lim_{b\to\infty}\int_{1}^{b}\frac{dx}{x^{\frac{3}{2}}+\frac{1}{2}}\leq\lim_{b\to\infty}2(1-\frac{1}{\sqrt{b}})=2$$
Okay with that one?

 

[arildno]

The significance of the above, is that the fact that the integral part is bounded between 0 and 2, is that this shows that it exists. If we are able to show that the evaluation at infinity (your last question) is finite as well, we have proven that the original comparison integral exists (is convergent), and thus, the sum is convergent (your goal).

Remember that $$ln(1+\epsilon)\approx\epsilon,\epsilon<<1$$
Use this to find the answer to your last question.

 

[saltydog]

Well guys, I think it would be interesting to determine under what constraints imposed upon f(n) does the following sum converge:

$$\sum_{n=1}^{\infty} ln(1+f(n))$$

Is there a class of functions which guarantees convergence? Is there some boundary such that:

$$|f(x)|<Mg(x)$$

for some g(x) and M which would guarantee convergence?

Just a though that's all. Obviously f(n) would need to be some kind of "decaying" function which tend to 0 at infinity. How fast does it need to do this to assure convergence? You know it's just me but Twoflower, if you're taking Analysis right now, I say you figure it out. As always, do as you wish.

Edit: Anyway, if you wish so, one "initial attempt" would be to work on 10 or 12 of them near the point where your best guess would suggest convergence is lost and just see what happens. I suppose if there's a nice analytical proof, that would work too. It's good though to "experiment" with test cases however.

 

[twoflower]

Thank you arildno, now I think I finally have it. You enriched my horizons, your ideas about bounding the integral wouldn't come to my mind.

Anyway, here's another solution I actually wrote in the test, but I didn't want to post it here since I wasn't sure that it was ok. But as we were given the test results few hours ago our professor also showed how it could have been solved.

So,

$$\log \left( 1 + \frac{1}{2\sqrt{n^3}}\right) \approx \frac{1}{2\sqrt{n^3}}$$

And thus
$$\sum_{n=1}^{\infty} \log \left( 1 + \frac{1}{2\sqrt{n^3}}\right) < \infty$$

 

[steven187]

hello all

well from my understanding in terms of integration, let f(x) be the intergrand of the integral if f(x) is continuous on the interval [1,infinity) then find a g(x) which is also continuous and non negative on [1,infinity) such that |f(x)|<=g(x) for all x an element of [1,infinity) if the integral of of g(x) exists then so does the integral of f(x) and if f(x) is integrable then the integral is finite which then implies that it must converge on the interval [1,infinity), so what you need to do is find g(x) to bound it by it could be 1/x or
1/(2*x^(3/2)) since they both behave in a similar maner to f(x), what you have above is similar to what i have explained except that it is in terms of series rather than integrals so that your aim is to find a comparision series that behaves in a similar manner

steven

 

[arildno]

Thank you arildno, now I think I finally have it. You enriched my horizons, your ideas about bounding the integral wouldn't come to my mind.

Anyway, here's another solution I actually wrote in the test, but I didn't want to post it here since I wasn't sure that it was ok. But as we were given the test results few hours ago our professor also showed how it could have been solved.

So,

$$\log \left( 1 + \frac{1}{2\sqrt{n^3}}\right) \approx \frac{1}{2\sqrt{n^3}}$$

And thus
$$\sum_{n=1}^{\infty} \log \left( 1 + \frac{1}{2\sqrt{n^3}}\right) < \infty$$

This is, I believe, what Saltydog hinted at.
I would say that this is an excellent heuristic way of arguing for the convergence; the convergence is, however, not rigorously proved by this argument.
I posted a way in which we may rigorously prove the the result, by use of the integral test.
I'm not denying that we might develop a rigorous proof from the observed asymptotic behaviour of the tail end of our series; that is redily developed by showing that you have an alternating and decreasing Taylor series expansion of each term in your series.

I don't think this is much simpler than the unproblematic application of the integral test, though.