Solving series with alternating sign

[Dell]

is there any other way of solving conditional convergence/divergence for a series with alternating sign, whose absolute values' series diverges?

i have a question where i need to solve the following series,


$$\sum$$(-1)^n-1$$\frac{1}{n - sin(n)}$$ from 1 to infinity

i take the absolute value of An, and get An=$$\frac{1}{n-sin(n)}$$ now i take Bn=1/n which i know diverges with the limits 1, infinity.

lim(n->inf) $$\frac{Bn}{An}$$=$$\frac{n-sin(n)}{n}$$=1 therefore An behaves like Bn ==> diverges

if i take the abs values of the whole series i get i series which diverges leaving me with only 2 options, divergence or conditional convergance.

from here the only way i know of i leibniz laws,
-lim An =0
-An>A(n+1)

for the limAn =0 exists, but as for An>A(n+1) i cannot prove this since An is dependant on sin(n) which can be anywhere between -1 and 1 and is not necessarily larger or smaller for n+1 than for n. because of this i would think that the series diverges, but this is not the case. how do i prove that the series converges??

 

[Russell Berty]

That is a sticky one.

Try this:

Rewrite your series by combining consecutive terms.

Notice that (-1)^(n-1)*1/(n - sin n) + (-1)^(n)*1/(n + 1 - sin(n+1)) can be written as a single fraction and the new denominator has a degree 2 poly in n whereas the numerator simplifies to a bounded value. Test for absolute convergence of this new series.

 

[Dell]

how can i just rewrite my series like that? surely this will NOT give me the same series??

 

[Dell]

is it true to say that for n->infinity, An=(-1)^(n-1)*1/n which i know converges, therefore An converges.?

 

[lanedance]

have you tried Russell Berty's idea? i think it will work

you are not really rewriting the series, just summing adjacent terms without changing the order...

if i remember right conditional convergance for the alternating series
$$\sum (-1)^n a_n$$
relies on showing a_n is monotone decreasing and tends to zero, then the alternating series converges

so
$$a_n = \frac{1}{n-sin(n)}$$
clearly tends to zero as n gets large

however its not so clear a_n is monotone decreasing, that would require
$$a_n - a_{n+1}>0$$ for all n
but
$$a_n - a_{n+1}= \frac{1}{n-sin(n)} - \frac{1}{n+1-sin(n+1)} = n+1-sin(n+1) -n+sin(n) = 1-sin(n+1) +sin(n)$$

this must be >0 can you show that?

 

[Dell]

i suppose that that would ork, but in another, very similar series -1^(n-1) * 1/(n+100sin(n)) it doesn't work anymore, how would you solve that then??/

also can leibniz be used if the series is not monotone decreasing for any an, (like in the case of -1^(n-1) * 1/(n+100sin(n)) ? if not how do i solve a series which is like that?

 

[lanedance]

in the last post i meant you would have to show it is monotone decreasing

i think the original suggestion might be the way to go for what you want...

 

[Dell]

in the original question it is ... +100sin(n) and not ...-sin(n) in which case i don't hink i can solve it in that way, but i don't think that the series is monotone decreasing so what should i do??


An= -1^(n-1) * 1/(n+100sin(n))

 

[Russell Berty]

You can use the trick i gave you for any series of the form

a(n) = (-1)^(n-1)/[n+b*sin(n)] where b is a constant (100, or 1, or whatever.)

Notice two facts:
1) The sum of two consecutive terms --->

a(2k-1) + a(2k) = 1/[2k-1+b*sin(2k-1)] + (-1)/[2k+b*sin(2k)]

= [2k + b*sin(2k) - (2k - 1 +b*sin(2k-1)]/ ( [2k-1+b*sin(2k-1)]*[2k+b*sin(2k)] )

= [1 + b*(sin(2k) - sin(2k-1))]/[ (2k -1)2k + (2k-1)b*sin(2k) + 2k*b*sin(2k-1) +b*b*sin(2k)*sin(2k+1) ]

= [1 + bounded value]/[4k^2 + bounded value*k + bounded value]

Use a limit comparison with 1/k^2 to see the series [a(2k-1)+a(2k)] converges.

Let L be the limit of this series.


Fact 2) a(n) -> 0 as n -> 0 (as you stated.)

From these two facts, we see that for any epsilon > 0, we can find an N such that
| the sum from 1 to m of [a(2k-1)+a(2k)] - L | < epsilon/2
for all m>N (by fact 1)
AND we can find an M such that |a(n)|<epsilon/2 for all n>M (by fact 2.)

Now let W = the maximum of N and M. Then for any n > W, The sum from 1 to n of
[a(2k-1)+a(2k)] is less than epsilon/2 AND |a(n)|<epsilon/2.

So for your alternating series we know: for any n > W

|a(1) + a(2) + a(3) + a(4) + ... + a(2n-1) + a(2n) - L| < epsilon/2

And also (using the triangular inequality)

|a(1) + a(2) + a(3) + a(4) + ... + a(2n-1) + a(2n) + a(2n+1) - L| <

|a(1) + a(2) + a(3) + a(4) + ... + a(2n-1) + a(2n) - L| + |a(2n+1)| < epsilon/2 + epsilon/2

= epsilon.

The idea is this: We know that the partial sums of your series converge if we look at only the sums of the first 2k terms (my series). For your series, we still need to consider the partial sums of the first 2k + 1 terms, which we notice is only different by a single term to the partial sum of the first 2k terms. This is where we use the fact that the extra term, the 2k+1 st term, gets arbitrarily small.

 

[Dell]

for the same doing, could i not just say that my series behaves like (-1)^(n-1)*1/n for very large n's, which is where i fear diverence, and since i know that (-1)^(n-1)*1/n converges, so will my series??

 

[Russell Berty]

That is the intuition behind the argument. The problem is describing in what way it "behaves like" (-1)^(n-1)/n. This series is conditionally convergent and those kinds of series are very touchy. Slight modifications (with a similar "behavior") can cause divergence.

As you pointed out, this series does differ from (-1)^(n-1)/n because the An's are not decreasing. And this causes a big difference in "behavior" when looking at a series that is not absolutely convergent.

The bottom line is: Saying it "behaves" like (-1)^(n-1)/n is not a sufficient explanation in a proof.