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Calculus 2 - Infinite Series

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  • #1
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State if the following series converges or diverges
sum[k=1,inf] cos(k)/(k^2+1)

I applied the convergence test
lim k->inf cos(k)/(k^2+1) =H lim k->inf -sin(k)/(2k) =H lim k->inf -cos(k)/2 = some undefined value bounded by -1 and 1 =/= 0 so the series diverges by the divergence test. I guess my answer is wrong.

This question came up on a assignment in class. We went over the answer after handing in the the assignment. The lecturer stated that the the series did not converge and compared it to sum[k=1,inf] 1/k^2 and stated that

0 < sum[k=1,inf] cos(k)/(k^2+1) < sum[k=1,inf] 1/k^2

and because sum[k=1,inf] 1/k^2 is convergent by the p-series test, 2>1, sum[k=1,inf] cos(k)/(k^2+1) must also converge by the comparison test sense sum[k=1,inf] cos(k)/(k^2+1) < sum[k=1,inf] 1/k^2.

I believed this to be true at the time this was stated. I however have a problem accepting this because
0 < sum[k=1,inf] cos(k)/(k^2+1)
is not true statement because cos(k) alternates between positive values and negative values.

I also have a problem with this statement
sum[k=1,inf] cos(k)/(k^2+1) < sum[k=1,inf] 1/k^2

cos(k) is bounded by positive and negative one and is undefined as k goes to infinity but stuck in between these two values. As k goes to infinity
k^2+1 = k^2
so the denominators would be equal to each other
and at some very large values cos(k) = 1
so wouldn't this statement be more correct
sum[k=1,inf] cos(k)/(k^2+1) <= sum[k=1,inf] 1/k^2
and I'm not sure the comparison test works here when you have >= or <= comparisons. I thought they had to be strictly < or >

Just from thinking about it... I would like to believe that we cannot say anything at all about the series with comparison tests with >= or <=... but I'm not so sure as to exactly why at the moment...

I checked with wolfram alpha
http://www.wolframalpha.com/input/?i=sum[n=1,+inf]+of+cos(k)/(k^2+1)"
and it says the series does not converge by the limit test as I showed. I however know computers can be wrong sometimes and have seen wolfram alpha return wrong results before, but I'm not so sure that the series converges and don't see any test that I know of that would be practical to use besides the comparison test or the limit comparison test in which it's hard to state < or > or <= or >= because cos(k) alternates between positive one and negative one and is undefined as k goes to infinity but and is stuck some were in between positive and negative one including those values.

Thanks for any help.
 
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  • #2
HallsofIvy
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State if the following series converges or diverges
sum[k=1,inf] cos(k)/(k^2+1)

I applied the convergence test
lim k->inf cos(k)/(k^2+1) =H lim k->inf -sin(k)/(2k) =H lim k->inf -cos(k)/2 = some undefined value bounded by -1 and 1 =/= 0 so the series diverges by the divergence test. I guess my answer is wrong.

This question came up on a assignment in class. We went over the answer after handing in the the assignment. The lecturer stated that the the series did not converge and compared it to sum[k=1,inf] 1/k^2 and stated that

0 < sum[k=1,inf] cos(k)/(k^2+1) < sum[k=1,inf] 1/k^2

and because sum[k=1,inf] 1/k^2 is convergent by the p-series test, 2>1, sum[k=1,inf] cos(k)/(k^2+1) must also converge by the comparison test sense sum[k=1,inf] cos(k)/(k^2+1) < sum[k=1,inf] 1/k^2.
So what did your lecturer say? You have said he "stated that the the series did not converge" and that he said it "must also converge by the comparison test".

I believed this to be true at the time this was stated. I however have a problem accepting this because
0 < sum[k=1,inf] cos(k)/(k^2+1)
is not true statement because cos(k) alternates between positive values and negative values.
Did you try calculating a few terms? If k= 1, cos(1)/(1+1)= 0.2702. If k= 1 cos(2)/(4+1)= -0.0832 so the sum is 0.1870, still positive. The denominator, [itex]k^2+ 1[/itex], decreases the value of the fraction fast enough that every partial sum is positive.

I also have a problem with this statement
sum[k=1,inf] cos(k)/(k^2+1) < sum[k=1,inf] 1/k^2

cos(k) is bounded by positive and negative one and is undefined as k goes to infinity but stuck in between these two values. As k goes to infinity
k^2+1 = k^2
No, that is not true. [itex]k^2+1> k^2[/itex] for all k. You cannot say "as k goes to infinity" and then use k in the formula. As k goes to infinity, the difference between [itex]k^2+1[/itex] and [itex]k[/itex] is always 1.

so the denominators would be equal to each other
and at some very large values cos(k) = 1
No, that is not true. cos(x)= 1 only for x a multiple of [itex]2\pi[/itex]. An integer, k, is never a multiple of the irrational number, [itex]2\pi[/itex].

so wouldn't this statement be more correct
sum[k=1,inf] cos(k)/(k^2+1) <= sum[k=1,inf] 1/k^2
and I'm not sure the comparison test works here when you have >= or <= comparisons. I thought they had to be strictly < or >
Well, if it is "=" and the sum on the right converges, the sum on the left must also converge!
No, it does not have to be "strictly < or >":
http://en.wikipedia.org/wiki/Convergent_series#Convergence_tests'

I checked with wolfram alpha
http://www.wolframalpha.com/input/?i=sum[n=1,+inf]+of+cos(k)/(k^2+1)"
and it says the series does not converge by the limit test as I showed. I however know computers can be wrong sometimes and have seen wolfram alpha return wrong results before, but I'm not so sure that the series converges and don't see any test that I know of that would be practical to use besides the comparison test or the limit comparison test in which it's hard to state < or > or <= or >= because cos(k) alternates between positive one and negative one and is undefined as k goes to infinity but and is stuck some were in between positive and negative one including those values.

Thanks for any help.
 
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  • #3
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4,991
State if the following series converges or diverges
sum[k=1,inf] cos(k)/(k^2+1)

I applied the convergence test
What you're applying is the divergence test (AKA nth term test for divergence).
lim k->inf cos(k)/(k^2+1) =H lim k->inf -sin(k)/(2k) =H lim k->inf -cos(k)/2 = some undefined value bounded by -1 and 1 =/= 0 so the series diverges by the divergence test. I guess my answer is wrong.
Your first mistake is using L'Hopital's Rule to evaluate the limit. L'Hopital's Rule can be used for limits that have the indeterminate form [0/0] or [±∞/∞]. Although k2 + 1 approaches 0 as k approaches ∞, cos(k) does not have a limit as k approaches ∞.

The expression has a limit that can be evaluated directly:

[tex]\lim_{k \to \infty} \frac{cos(k)}{k^2+1} = 0[/tex]

If you're using the nth term test for divergence, this fact does you no good. You can conclude that a series diverges only if lim an [itex]\neq[/itex] 0.
 
  • #4
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So what did your lecturer say? You have said he "stated that the the series did not converge" and that he said it "must also converge by the comparison test".
Well he said that the answer was that it converges. Sorry if I got that mixed up. He used

0 < sum[k=1,inf] cos(k)/(k^2+1) < sum[k=1,inf] 1/k^2

and because sum[k=1,inf] 1/k^2 is convergent by the p-series test, 2>1, sum[k=1,inf] cos(k)/(k^2+1) must also converge by the comparison test sense sum[k=1,inf] cos(k)/(k^2+1) < sum[k=1,inf] 1/k^2.

Did you try calculating a few terms? If k= 1, cos(1)/(1+1)= 0.2702. If k= 1 cos(2)/(4+1)= -0.0832 so the sum is 0.1870, still positive. The denominator, k^2+1, decreases the value of the fraction fast enough that every partial sum is positive.
I did not realize this at the time. The numerator is always strictly less than 1 and strictly greater than -1 and the denominator increases with each term. Can we say though that every partial sum is always positive? I don't see how we can.

No, that is not true. k^2+1>k^2 for all k. You cannot say "as k goes to infinity" and then use k in the formula. As k goes to infinity, the difference between k^2+1 and k is always 1.
Interesting. I thought the difference of 1 was irrelevant as k goes to infinity and that we could consider them to be about the same value.

Your first mistake is using L'Hopital's Rule to evaluate the limit. L'Hopital's Rule can be used for limits that have the indeterminate form [0/0] or [±∞/∞]. Although k2 + 1 approaches 0 as k approaches ∞, cos(k) does not have a limit as k approaches ∞.
Right because cos(x) is bounded by +/- 1 and so L'Hopital's rule can not be used. I didn't realize I made this mistake.


So does the series converge or diverge? Wolfram Alpha says that the sum diverges.
 
  • #5
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[tex]\frac{-1}{k^2+1} \leq \frac{cos(k)}{k^2+1} \leq \frac{1}{k^2+1}[/tex]

Also, since [itex]\frac{1}{k^2 + 1} < \frac{1}{k^2}[/itex], we can say that [itex]\sum_{k=1}^{\infty}\frac{1}{k^2+1}[/itex] converges. A similar argument shows that [itex]\sum_{k=1}^{\infty}\frac{-1}{k^2+1}[/itex] converges.

What can you conclude from this and your instructor's explanation? I don't trust what you're reporting that wolframalpha found, since I don't know what you entered there.
 
  • #6
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Interesting. So the series converges by the squeeze therom. This explanation is much more convincing.

I entered this into wolfram alpha
sum[n=1, inf] of cos(k)/(k^2+1)
and it said that the sum does not converge.
 
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  • #7
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I followed your link to wolframalpha to see for myself what you had done.
Your link shows this expression: sum[n=1, inf] of cos(k)/(k^2+1)

The expression should actually be sum[k=1, inf] of cos(k)/(k^2+1).

After fixed that, I got a slightly different result, but still one that said the series doesn't converge. What I have in post #5 doesn't guarantee that the series converges, only that the sequence of partial sums is bounded below by whatever [itex]\frac{-1}{k^2+1}[/itex] converges to (about -1.08), and above by whatever [itex]\frac{1}{k^2+1}[/itex] converges to. That's not enough to guarantee the the series in this thread converges. Wolframalpha says that the series diverges by the integral test, but I don't think it's trivial to evaluate that integral.
 
  • #8
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Your link shows this expression: sum[n=1, inf] of cos(k)/(k^2+1)

The expression should actually be sum[k=1, inf] of cos(k)/(k^2+1).
Ah thanks for pointing out my error.

After fixed that, I got a slightly different result, but still one that said the series doesn't converge. What I have in post #5 doesn't guarantee that the series converges, only that the sequence of partial sums is bounded below by whatever −1k2+1 converges to (about -1.08), and above by whatever 1k2+1 converges to. That's not enough to guarantee the the series in this thread converges. Wolframalpha says that the series diverges by the integral test, but I don't think it's trivial to evaluate that integral.
Oh that makes sense because -1/(k^2+1) and 1/(k^2+1) converge to different values, not the same value, so the series must be bounded between the two values but does not necessarily converge.

I thought about applying the integral test when I was doing this assignment and like you said got stuck and thought it wasn't possible. I saw the term
k^2 + 1
and automatically thought of doing a trig substitution, which would work fine but you would have to substitute k into cosine... hmm... I can't think of any way to solve this problem :cry:
 
  • #9
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Well by trig substitution...

integral cos(k)/(k^2+1) dk = - integral cos( cos(theta)/sin(theta) ) dtheta

I'm stumped.
 
  • #10
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Yeah, I don't have any ideas, either. The best I can offer is to think about all of the tests you could possibly use, and see if there's one that applies here.
 
  • #11
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  • #12
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One criterion for the integral test is that the integrand must be nonnegative. The function you're integrating doesn't satisfy that condition.
 
  • #13
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One criterion for the integral test is that the integrand must be nonnegative. The function you're integrating doesn't satisfy that condition.
Right I forgot about this. So how is wolfram alpha able to say by the integral test the series is does not converge than?
 
  • #14
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could I separate the integral?

integral cos(k)/(k^2+1) dk = integral cos(k) dk * integral 1/(k^2+1) dk

or is this illegal?
 
  • #15
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No you can't do that. This would be like saying that
[tex]\int x^2 dx = \int x dx \cdot \int x dx[/tex]

As far as WolframAlpha being able to integrate it, they're using some techniques that I don't know about.
 
  • #16
Ray Vickson
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If t(k) = cos(k)/(1 + k^2), we have |t(k)| < 1/k^2, so the series sum_k t(k) is absolutely convergent.

RGV
 
  • #17
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Right I was thinking multivariable
why does wolfram alpha say that the sum does not converge by the integral test?
 

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