Interval of convergence

Question

a.) Evaluate ∫cosx/x dx. b.) Give the interval of convergence of the series in part a.

I've gotten ∫cosx/x = Ci(x)+C so far

Answer 1

Hey! This one is a bit of a tricky one- you cannot just plug it into an online calculator as you tried. The answer you provide, ' Ci(x) + C' effectively means 'the integral of a cosine,' which is not very helpful. Instead, we have to use series.

a)

In class, you likely learned about the Taylor Series about a=0, known as a Maclaurin series. We must use the Maclaurin expansion of cos(x) to help us solve this problem. (If you do not remember what this is, I suggest looking it up to refresh your memory or asking about it again in this thread if you do not understand it).

Executing a Maclaurin expansion on cos(x) yields: cos(x) ≈ (1 - x²/2! + x⁴/4! - x⁶/6! ...)

If we plug this approximation into the integral for cos(x), we get:

∫ (1 - x²/2! + x⁴/4! - x⁶/6! ...)/x dx

or, by factoring the 1/x into the series,

∫ (1/x - x/2! + x³/4! - x⁵/6! ...) dx

which is something much easier to integrate! We can integrate each term one by one to get

ln(|x|) - (1/2)(x²/2!) + (1/4)(x⁴/4!) - (1/6)(x⁶/6!) ... + C (don't forget the constant, as it is an indefinite integral).

Aside from the ln(|x|) term, this can be rewritten as the series starting at n=1 for ∞ terms:

ln(|x|) + ( ∑ (-1)ⁿ • (x²ⁿ) / [(2n) • (2n)!] ) + C

This should be the answer to part a, and I hope my explanation helps you understand the process.

b)

There are a couple ways to find the interval of convergence, but I typically advise you utilize the ratio test first, especially in this case and you may see why later.

The ratio test says if we can define a series by each term a_n as we did at the end of part a, then we can define

L = lim_n→∞|(a_n+1) / (a_n)|

and if L < 1, then the series is convergent everywhere and if L > 1 then the series is divergent and thus the interval DNE.

As part a is the more difficult and tricky part of this problem, I will leave the rest of part b for you to solve, but if you do need further help or if you find a mistake in my work, please do follow up. I hope this helps!

EDIT TO HELP MORE WITH PART B)

Without giving away the answer, 'n' typically represents what term you are on (for example, n=1 means the first term in a series, n = 2 means the second, and so on. Though, technically it is simply what number you plug in for that term, so it can start at the zeroth term).

Thus, a_n+1 is the next term in the series. If your series was defined as the series of a_n = n starting at n = 0, then your series would go:

0, 1, 2, 3, 4 ....

So, for the ratio test, you plug in what that next term (a_n+1) would be- except you can't know what term you are at, so you have to generalize it for all terms. Thus, for the ratio test, you plug in the generalized next term over the generalized current term. I understand that phrasing is convoluted, so to put it in the necessary steps of the ratio test would be

Find the generalized next term by plugging in "(n+1)" into your series.

Find the generalized current term by plugging in "n" into your series.

Divide the first step result over the second step result.

Take absolute value of that.

Take limit of that as n goes to infinity.

This is always true. So, if your series was instead the sum of a_n = n, then the ratio test eqn would be:

L = limn→∞|(n+1) / (n)|

Or if each term was a_n = n^2, then it would be

L = limn→∞|(n+1)^2 / (n)^2|

Short answer: The ratio test effectively takes the absolute value of the ratio of a term over the term before it to see if it converges. If it is less than 1, this means the terms are getting smaller, so its sum eventually will converge.

To do the ratio test, plug in "(n+1)" wherever you see 'n' in the equation. This will be the numerator in the ratio test, and it is defined as "a_n+1" aka the formula for the next term.

Then, plug in "n" wherever you see "n" (yes, it should be exactly the same). The result will be your denominator, aka the current term in the series.

Now, divide the numerator by the denominator, take the absolute value of that, and take the limit as n→∞. You should be able to cancel things out using exponent rules or factoring and perhaps use L'Hopitals rule sometimes. What used to help me in class is changing 'n' and 'n+1' to 'x' and 'x=1' respectively, because it helped me know I could cancel things out.

If the limit of the absolute value of the ratio (numerator divided by denominator) is less than 1, then it converges everywhere and its interval is (-∞,∞), and if it is greater than 1, then it never converges and its interval DNE. Hope this helps! Feel free to comment again for further help or if you need perhaps a shorter explanation, haha.