Integral Test: Analyzing $$\sum_{n=3}^{\infty} \frac{1}{n^2-4}$$

[Yggdresil]

Homework Statement

Use the integral test to determine whether $$\sum_{n=3}^{\infty} \frac{1}{n^2-4}$$ converges or diverges.

The Attempt at a Solution

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Taking the integral we have $$\int_{}^{\infty}\frac{dn}{n^2-4}$$ Note: Mary Baos text is having us write the integrals without lower bounds to avoid errors in the evaluation. Weird, but it is what it is. $$-\frac{1}{4}\int_{}^{\infty}\frac{dn}{(\frac{-n}{2})^2+1}$$ $$-\frac{1}{2}\int_{}^{\infty}\frac{du}{-u^2+1}$$ which evaluates to $$-\frac{1}{2}\tanh^{-1}({\frac{n}{2}})\big|^{\infty}$$ Taking the limit $$-\frac{1}{2}\tanh^{-1}({\lim_{n \to \infty}\frac{n}{2}})$$It looks like the real component will go to zero, while the non real component does not. So my question is: for the integral test does it matter if there is some nonreal component to the limit? Or will this series converge because the real component tends to zero? Is the sub of arctanh valid?

I think I did the integral right, but I could have made a mistake as well.

 

[fresh_42]

I think it is better to use the logarithm than $\operatorname{arctanh}$ because of its domain. I suggest therefore to substitute $u :=\dfrac{2-x}{2+x}$ which allows you to deal with the limit to infinity. And I definitely recommend to write the lower bound $x=3\,.$

 

[Yggdresil]

I think it is better to use the logarithm than $\operatorname{arctanh}$ because of its domain. I suggest therefore to substitute $u :=\dfrac{2-x}{2+x}$ which allows you to deal with the limit to infinity. And I definitely recommend to write the lower bound $x=3\,.$

Thanks for the help, I’ll give that a go.

 

[vela]

Is the sub of arctanh valid?

You used the substitution $n = 2\tanh \theta$, but this isn't a good choice because $-2 < \tanh \theta < 2$ for real $\theta$ while $n\ge 3$ in the integral. You could use the substitution $n = 2\coth \theta$ instead, or use partial fractions.

 

[Yggdresil]

You used the substitution $n = 2\tanh \theta$, but this isn't a good choice because $-2 < \tanh \theta < 2$ for real $\theta$ while $n\ge 3$ in the integral. You could use the substitution $n = 2\coth \theta$ instead, or use partial fractions.

You’re right of course, I misspoke. Doesn’t the sub of $n = 2\coth \theta$ produce $-\frac{1}{2}\coth^{-1}(\frac{2}{n})$ which still has an imaginary component?

Using Fresh_42’s or your recommendation of partial fractions I can solve it in terms of logs $\frac{1}{4}(\ln(4n-8)-\ln(4n+8))\big|_{3}^{\infty}$ which when evaluated at the upper limit produces zero without an imaginary component. If I ran through the math right and didn’t miss anything.

 

[vela]

You’re right of course, I misspoke. Doesn’t the sub of n=2cothθn=2coth⁡θn = 2\coth \theta produce −12coth−1(2n)−12coth−1⁡(2n)-\frac{1}{2}\coth^{-1}(\frac{2}{n}) which still has an imaginary component?

You should get $n/2$ in the argument of the arccoth and a purely real result.

 

[Yggdresil]

You should get $n/2$ in the argument of the arccoth and a purely real result.

Ohh, let me run through it again. Thanks!

 

[Ray Vickson]

Ohh, let me run through it again. Thanks!

It is easiest to write
$$\frac{4}{u^2-4} = \frac{1}{u-2} - \frac{1}{u+2}$$ and then integrate each term separately.

It is a good strategy to try such "partial fractions" first and then use a change-of-variable technique if the partial-fraction method fails (or involves complex numbers, which start to become inconvenient).

 

[mathwonk]

i suggest noticing first this series converges if the simpler one with terms 1/n^2 does. i.e. after a while n^2-4 is larger t han n^2/2, hence the reciprocal 1/(n^-4) is eventually smaller than 2/n^2. this is much easier than those complicated substitutions, since now the integral test is trivial.

 

[alan2]

It’s a straightforward trig substitution. As for dropping the lower limit, that’s because only the tail of the series determines convergence, assuming that the beginning of the series is not infinite.

 

[Yggdresil]

It is easiest to write
$$\frac{4}{u^2-4} = \frac{1}{u-2} - \frac{1}{u+2}$$ and then integrate each term separately.

It is a good strategy to try such "partial fractions" first and then use a change-of-variable technique if the partial-fraction method fails (or involves complex numbers, which start to become inconvenient).

That’s a good point, I saw something that looked like a trig substitution , and got blinded by that route.

i suggest noticing first this series converges if the simpler one with terms 1/n^2 does. i.e. after a while n^2-4 is larger t han n^2/2, hence the reciprocal 1/(n^-4) is eventually smaller than 2/n^2. this is much easier than those complicated substitutions, since now the integral test is trivial.

Yeah, however it was just one of those exercises that explicitly stated the method to be used. I’m not sure if using a comparison would have been following “the spirit of the problem”, but I agree with you.

It’s a straightforward trig substitution. As for dropping the lower limit, that’s because only the tail of the series determines convergence, assuming that the beginning of the series is not infinite.

If you see a “straightforward” trig substitution I don’t! Beyond the result of the integral I was mostly curious about imaginary portions of my original result, which was resolved when it was pointed out earlier that I used a sub that wasn’t the best choice.

 

[mathwonk]

if the spirit of the problem was to introduce the integral test as a first step instead of the last step, then one could give a trig substitution but then also point out that this problem is more easily done by reducing to the case of 1/n ^2, and then using the integral test on that series. that way you satisfy all literal requirements of the problem and also show that you have some insight.

moreover this method also allows one to immediately show, by completing the square, that all series of form SUM 1/quadratic, are convergent. by comparing all of them with 1/n^2. Of course you could presumably also do this strictly by the integral test, since to do the integral you would first complete the square and then make a trig substitution. However why not point out that by doing the completion of the square first, and thus simplifying both the series and the integral, life is much easier.

Well I guess it matters what one is trying to teach by the problem. If one wants to enforce practice in doing trig substitutions then perhaps one does not want to make that substitution unnecessary. But if one just wants to understand why the series is convergent, i recommend making an initial comparison. I.e. if you have this insight you realize that any series of form SUM 1/quadratic, converges simply because the denominator grows quadratically, i.e. it converges because 1/n^2 does so, which in turn converges by the integral test.

I confess that although I believe I can see the fact that SUM 1/quadratic converges by comparison with 1/n^2, in my head, I do not see it by the integral test without actually working out the trig integral to see what it gives. Of course everyone has a different notion of what is "easy" or "intuitive".By the way, if one is asked whether the integral of 1/(x^2+1) converges, what if one remarks that it does so because it is smaller than the integral of 1/x^2? Does that satisfy the requirement to use the integral test? I.e. the integral test says the series converges if the integral does so. But does it require you to compute the exact integral, or just argue that the integral does converge?

I.e. what about saying that the integral of 1/(n^2-4) must converge because it is eventually smaller than the integral of 2/n^2, which is easily computed. Hence the series SUM 1/(n^2-4) converges.

Again I would argue that "use the integral test" does not mean one must use it in an inefficient way.

I guess after all this, what I am trying to argue is that proof of convergence by the integral test is a type of comparison test, and never yields the exact sum of the series, only the fact that that sum is finite. Hence there is no need and indeed no reason to compute exactly the same integrand as the term of the given series, as it still only gives you an approximation, i.e. a comparison. All one wants to know is whether or not the integral is finite, and this never need involve computing the precise integral. I.e. I claim one should understand that computing the exact integral of the given term in the series, does little more good than computing a much simpler comparison integral. I.e. the value of the integral does give one an upper bound for the series, but so does the value of a simpler integral, although maybe not quite as good a bound.

Thus I suggest the only time one should actually compute the integral without simplifying it, is when one wants to use the value of the upper bound given by the integral and wants it to be as small as possible.

I beg pardon for the long essay if it does not help. (At least it helped me, i.e. I think I persuaded myself, and now feel I understand the use of the integral test better than I did before.)

OK: by the same method one gets that (if we sum only from one more than the largest root of the polynomial), that SUM 1/(polynomial in n) is also convergent as soon as the polynomial has degree ≥ 2.

I.e. writing a polynomial of form x^n + ...+a = x^n (1+...+a/x^n), we see that for lartge n, the second factor is closer to 1 than 1/2, and hence the product is larger than 2/x^n. Hence we can either compare the series to this series, or we can compare the resulting integral to the integral of 2/x^n, for n ≥ 3, which is thus convergent. (To satisfy the monotonicity hypothesis of the integral test, note that a non constant polynomial is not only eventually non zero, but also is eventually monotone, since its derivative is eventually non zero.)

Obviously however no one can evaluate all these integrals of form 1/polynomial, exactly. Hence prudent use of the integral test, i.e. by comparison, yields vastly better results than rigid insistence on exact integral evaluations.

Remark: Beginners in partial fractions might suggest that one can use that technique to evaluate all such integrals exactly, but experts know well that this is hopeless since one cannot in practice factor most polynomials, which is the first step.