Question

Does the series converge or diverge? If it converges, find its sum: Sum of (1/(n^3-n)) from 2 to infinity Does the series converge or diverge? If it converges, find its sum: Sum of (1/(n^3-n)) from 2 to infinity @Mathematics

Answer 1

It is a convergent series.

Answer 2

I tried doing partial fractions and ended up with up
∑(−1/n+1/(2(n+1)+1/(2(n−1))
I know the 1/n part diverges since it's a harmonic series. I also tried the integral test on the 2 other terms but they both tend to infinity. But Wolfram alpha says the sum converges to 1/4... I dunno how find it :(

Answer 3

maybe to make it clear. i have:
-1/n + 1/(2(n+1) + 1/(2(n-1))

Answer 4

You can play with these indices to make it clear. Notice that
\[- \sum_2^\infty \frac{1}{n} + \frac{1}{2}\sum_2^\infty \frac{1}{n+1} + \frac{1}{2}\sum_2^\infty \frac{1}{n-1}\]
can also be written as
\[- \sum_2^\infty \frac{1}{n} + \frac{1}{2}\sum_3^\infty \frac{1}{n} + \frac{1}{2}\sum_1^\infty \frac{1}{n}\]
So, if we pull the n=2 term from the first series and both the n=1 and n=2 terms from the second series, we get....
\[\sum_3^\infty (\frac{-1}{n} + \frac{1}{2}\frac{1}{n} + \frac{1}{2}\frac{1}{n}) -\frac{1}{2} + \frac{1}{2} + \frac{1}{4} = 0+\frac{1}{4} = \frac{1}{4}\]

Answer 5

Oh my :o thank you so much. I'm just ccurious though why my method made it seem divergent? 1/n is divergent as a harmonic series and 1/(2(n+1)) by the integral test would give 1/2ln|x+1| from 2 to inf which would be divergent o_O I find this sum strange :( Maybe one of my steps is wrong

Answer 6

No, in and of themselves they're divergent series, but together the divergences cancel each other out in the same way that if you took
\[\sum 1 \]
and \[\sum -1\]
both of those series, diverges, while their sum converges to zero.

Answer 7

oh :o interesting.. So I guess you can't do like with integrals; when you split up an integral into 2 integrals and if one of those diverges, then the whole integral diverges..

Answer 8

That's not necessarily true either. Note that
\[ \int_{-\infty}^{\infty}x dx = \int_{-\infty}^{0}xdx + \int_{0}^{\infty}xdx = 0\]
even though both integrals diverge when considered separately.

Answer 9

hmm.. I have a definition of improper integrals in my book where it says if either of the improper integrals on the right diverges, then the integral on the left diverges o_o

Answer 10

Google something called the Cauchy Principle Value of an integral.

Answer 11

haha Principal***

Answer 12

It can be thought of more or less as some kind of l'Hospital's Rule for integrals. Not so much in the practical sense, but it gives definition to integrals that are not defined in the traditional sense.