Series question that I am confused on why my method doesn't work

[indefiniteintegral]

So I'm currently, taking Calculus 2 and I'm on Series right now and in 11.2 it's the first chapter on series, and covers the Divergence test,
In the first set of problems it wants you to find if the series in convergent, and if so provide the sum

So here is the problem I was solving
[math]\displaystyle\sum_{i=2}^n\frac{2}{n^2-1}[/math]
And I see that if I take the limit of the series at infinity it goes to zero, which means its inconclusive as to if it converges or not.
So another test must be used.

So me being bored this weekend learned the integral test before any other test, and when I saw this problem , I thought just my luck!
I will use partial fractions, solve it as an improper integral, get the limit and I'm onto the next problem hooray for me easy peasy

So I set my f(x) as follow:

[math]\int^∞_2 \frac{2}{x^2-1}[/math]
[imath]=\int^∞_2 \frac{2}{x^2-1}= \int^∞_2 \frac{A}{x-1} +\frac{B}{x+1}[/imath]

[imath]=\int^∞_2 \frac{1}{x-1} -\frac{1}{x+1}[/imath]

[imath]= ln| \frac{x-1}{x+1}|[/imath]

So now I can take this and evaluate the limit from 2 to t as t goes to infinity

Which then leaves me with

[imath]= 0 - ln \frac{1}{3}[/imath]
[imath]= ln|3|[/imath]

So there was my final answer. But I looked it up in the textbook, and they "Used the definition of a convergent series and used partial fractions to get the partial sums"

They used [imath]\displaystyle\sum_{i=2}^n\frac{1}{i-1} - \frac{1}{i+1}[/imath]

From there they used a telescoping series to get a final answer which is not the answer I got.


So I'm sorry for typing so much, and I apologize if my question is stupid but my question is, if the sum of the series has to be calculated the way specifically done by the book, then what on earth did I calculate by using the integral test? Did I not calculate the sum of the function with the integral test? Is the sum of the function not the same as the sum of the series? I feel like I am fundamentally missing something here.

Thanks for any help!

 

[blamocur]

Replacing series with integrals often helps you to figure out whether the series converges, but the series' limit does not have to be the same as the value of the definite integral. For example[imath]\sum_{k=1}^\infty \frac{1}{2^k}=1[/imath], but, if I did my integral correctly, [imath]\int_1^\infty \frac{1}{2^x}dx = \frac{1}{2\ln 2} \approx 0.72135[/imath]

 

[indefiniteintegral]

Thank you for your insight, so if I understand you correctly, when I use the integral test, it will tell me if it's converges and if it does so, will give me an approximation of the limit? Or is the improper integral the more precise number? My mind is both confused and blown at the same time. Thanks for the help!!

 

[Steven G]

Please re-read what you read about the integral test. I assure you that it never said that if the integral converges than the series converges to that value

 

[Steven G]

I am thrilled that when you were bored that you decided to do Calculus. Excellent choice!

 

[indefiniteintegral]

Please re-read what you read about the integral test. I assure you that it never said that if the integral converges than the series converges to that value

I will! I really never read anything on the integral test, I had just happen to have a video on it in my recommendations on youtube, but I will read and study it properly today! I just assumed these values would converge at the same number, not sure why. I was so baffled that I came right here and asked my question haha sorry for my impatience!! But I am enjoying Cal 2 a lot, it challenges me, so I will keep up my excellent choice as you put it

 

[Dr.Peterson]

my question is, if the sum of the series has to be calculated the way specifically done by the book, then what on earth did I calculate by using the integral test? Did I not calculate the sum of the function with the integral test? Is the sum of the function not the same as the sum of the series?

if I understand you correctly, when I use the integral test, it will tell me if it's converges and if it does so, will give me an approximation of the limit? Or is the improper integral the more precise number?

It happens that I tutored two students today face-to-face who were doing the same homework on infinite series and the integral test, and one question asked about using the integral test to estimate the value of the sum. The method for this turned out to be hidden in the proof they were given for the integral test: The integral is an upper bound (overestimate) for the actual sum excluding the first term.

So in your problem, the first term, 2/3, plus the integral, ln(3) = 1.0986..., tells us that the sum is less than 1.765... . If the exact sum they found was, as I expect, 1.5, this illustrates the theorem, and shows what the integral test really does. It's not a very close estimate (and can't be easily improved), but it does tell you something about the value.

We normally use the integral test only to prove convergence or divergence, so when you are asked to find the value, this shouldn't be the main thing you do -- they want an exact value, not an upper bound. But when they don't ask for a value, the integral test is appropriate.