Infinite Sequences and Integrals

[Helicobacter]

I wrote everything on the scanned image:

http://img516.imageshack.us/img516/79/phprollyypmkz6.jpg

The solution to the first sequence problem is 0, i.e. it converges, which is a puzzle to me...

Ignore the limits in the second problem, all I need to know is how to integrate it.

Thanks for any help

 

[sutupidmath]

for the series, we know that if the limit of a_n of any given series as n -> infinity, is different from zero than the given series diverges, right?
so
lim n->infinity [(-1)^(n+1)]/(2n-1), first let's suppose that n is pair so n=2k, k is an integer. Now as you can see the limit tends to infinity.
On the other hand, if we suppose that n is odd, so let n=(2k+1), than again the limit tends to infinity. so we can conclude that the sereies diverges. There are other ways of proving this also.

 

[sutupidmath]

as for the integral i think that it is a tabelar integral. The indefinit integral of
integ dx/(x^2 - 1) = ln I(x-1)/(x+1), so here we just have the bounds, which go from c to -2. Also remember that integ of f(x) from a to b, is:
f(b)-f(a).

 

[d_leet]

for the series, we know that if the limit of a_n of any given series as n -> infinity, is different from zero than the given series diverges, right?
so
lim n->infinity [(-1)^(n+1)]/(2n-1), first let's suppose that n is pair so n=2k, k is an integer. Now as you can see the limit tends to infinity.
On the other hand, if we suppose that n is odd, so let n=(2k+1), than again the limit tends to infinity. so we can conclude that the sereies diverges. There are other ways of proving this also.

For one thing the original poster stated this was a sequence, not a series. For a second thing, regardless of how you look at it this sequence obviously converges, it has an upper bound of 1/(2n-1), qand a lower bound of (-1)/(2n-1) both of which converge to 0 taking the limit as n approaches infinity, hence the sequence in question converges using the squeeze theorem. Furthermore even if it were a series it would still converge by the alternating series test.

In response to the original poster, you cannot apply L'Hopital's rule as you have done because it is not of either of the forms 0/0 or infinity/infinity.

 

[sutupidmath]

For one thing the original poster stated this was a sequence, not a series. For a second thing, regardless of how you look at it this sequence obviously converges, it has an upper bound of 1/(2n-1), qand a lower bound of (-1)/(2n-1) both of which converge to 0 taking the limit as n approaches infinity, hence the sequence in question converges using the squeeze theorem. Furthermore even if it were a series it would still converge by the alternating series test.

In response to the original poster, you cannot apply L'Hopital's rule as you have done because it is not of either of the forms 0/0 or infinity/infinity.

Yeah, you are so right. My bad, i appologize for my wrong reply. I did not read it properly, just glanced at it. Sorry!

 

[VietDao29]

as for the integral i think that it is a tabelar integral. The indefinit integral of
integ dx/(x^2 - 1) = ln I(x-1)/(x+1), so here we just have the bounds, which go from c to -2. Also remember that integ of f(x) from a to b, is:
f(b)-f(a).

Well, no, it's definitely not f(b) - f(a). It should have read: F(b) - F(a), where F(x) is the anti-derivative of f(x).

Btw, are you sure that:
$$\int \frac{dx}{x ^ 2 - 1} = \ln \left| \frac{x - 1}{x + 1} \right| + C$$

You seem to have missed a factor of 1/2 before the final result. Actually, it should read:
$$\int \frac{dx}{x ^ 2 - 1} = \textcolor{red}{\frac{1}{2}} \ln \left| \frac{x - 1}{x + 1} \right| + C$$

:)

-------------

@OP:
You don't have to split the limit to 2 separate limits like that. Here's one way you can tackle the problem:
$$\lim_{c \rightarrow -\infty} \int_c ^ {-2} \frac{2 dx}{x ^ 2 - 1} = \lim_{c \rightarrow -\infty} \int_c ^ {-2} \frac{(x + 1) - (x - 1)}{(x - 1) (x + 1)} dx$$

$$= \lim_{c \rightarrow -\infty} \left[ \int_c ^ {-2} \left( \frac{1}{x - 1} - \frac{1}{x + 1} \right) dx \right] = \lim_{c \rightarrow -\infty} \left( \left. \ln \left| \frac{x - 1}{x + 1} \right| \right|_c ^ {-2} \right)$$

$$=\ln \left| \frac{(-2) - 1}{(-2) + 1} \right| - \lim_{c \rightarrow -\infty } \ln \left| \frac{c - 1}{c + 1} \right| = ...$$

You can take it from here, right? :)

 

[CompuChip]

As for the series... you could try dividing both the numerator and denominator by n and see if you can find the limit.

If you need to sum over all the a_n and show that this converges, it will be sufficient to prove absolute convergence, as
$$\sum a_n \le \left| \sum a_n \right| \le \sum |a_n|$$, so if the rightmost converges, so will definitely the leftmost.[/color] (never mind that, of course it doesn't converge absolutely) Now it's for you to show that not only do the terms drop to zero, the series also converges (this is not trivial, think about 1/n). (It's not hard by the way, if you have this criterion :) )