Testing convergence of a series

In summary: Anyway, a proof of the convergence of \sum_{n=1}^\infty \frac{\cos n}{n^2} by Abel's test:Let f(x) = \sum_{n=1}^\infty \frac{\cos n}{n^2} x^n. Then f is continuous on (-1,1), so we can consider the Abel sum \lim_{t\to 1^-} f(t).To see that this limit exists, we note that f is uniformly convergent on [0,1), since \sum_{n=1}^\infty \frac{1}{n^2} converges. This is because|\sum_{n=N}^M \
  • #1
MaximumTaco
45
0
[tex]
\sum_{n=1}^\infty \frac{cos n}{n^2}
[/tex]

Now, which test should i try? It doesn't appear to be anything straightforward like ratio test, comparison test etc and it's not a straightforward limit.
cos(n) goes negative too, so that complicates things.

Can anyone offer me any advice on a problem like this? Thanks!
 
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  • #2
Obviously this series has both negative and positive terms, but it's not an alternate series. The signs change is irregular.

If i were you i would the uniform convergence test.

[tex] \sum_{n=1}^{\infty} |\frac{cos n}{n^2}| = \sum_{n=1}^{\infty} \frac{|cos n|}{n^2} [/tex]

then i would try a comparison with a p-series such as [itex] \sum_{n=1}^{\infty} \frac{1}{n^2} [/itex]
 
  • #3
The convergence of your series follows from http://www.shu.edu/projects/reals/numser/t_abel.html .
 
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  • #4
maybe this is correct I am not sure

define a sequence [tex]An=\frac{\cos n}{n}[/tex]

this sequence converges to 0
[tex]\lim_{n \to \infty} An = 0[/tex]

[tex]\frac{A1+A2+...+An}{n}=\sum_{n=1}^n \frac{cos n}{n^2}[/tex]
[tex]\frac{A1+A2+...+An}{n}[/tex] is the arithmetical average of the sequence [tex]An[/tex]
and it always converges to the same limit as the sequence [tex]An[/tex]
so:
[tex]\lim_{n \to \infty} An = \lim_{n \to \infty} \frac{A1+A2+...+An}{n} = \sum_{n=1}^\infty \frac{cos n}{n^2}=0[/tex]
 
  • #5
Muzza said:
The convergence of your series follows from http://www.shu.edu/projects/reals/numser/t_abel.html .

Pretty conclusive Muzza. Thanks for the link to the Abel's test. :smile:
 
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  • #6
and it always converges to the same limit as the sequence
so:

IF the respective limits exist.

Consider the sequence: 0, 1, 0, 1, 0, 1, ...

The arithmetic mean converges to 1/2, but the sequence clearly does not.
 
  • #7
Anzas said:
[tex]\frac{A1+A2+...+An}{n}=\sum_{n=1}^n \frac{cos n}{n^2}[/tex]


That is wrong, try n=2:

(cos(1)+cos(2)/2)/2

which is the LHS, is not equal to

cos(1)+cos(2)/4

You are forgetting n is just the variable of the summation, when you're using it as the upper limit as well.

We could, if this were true, apply it with 1 in the numerator, instead, and you're claiming that the sum of 1/n^2 is zero, when it is clearly not.
 
  • #8
your right i didn't notice that.
o well just ignore my post then :smile:
 
  • #9
MaximumTaco said:
[tex]
\sum_{n=1}^\infty \frac{cos n}{n^2}
[/tex]

Now, which test should i try? It doesn't appear to be anything straightforward like ratio test, comparison test etc and it's not a straightforward limit.
cos(n) goes negative too, so that complicates things.

Can anyone offer me any advice on a problem like this? Thanks!

Let's try for absolute convergence.
[tex]
\sum_{n=1}^\infty abs(\frac{cos n}{n^2})
[/tex]
is smaller than (comparison test)
[tex]
\sum_{n=1}^\infty abs(\frac{1}{n^2})
[/tex]
which equals
[tex]
\sum_{n=1}^\infty \frac{1}{n^2}
[/tex]
which converges by a p-series test.
 
  • #10
WHOOPS! alketran beat me to it. never mind... :blushing:
 
  • #11
as courant makes so clear, almost the only way to prove convergence (except for the very special alternating test) is to prove absolute convergence by comparison.

there are essentially only two things to compare with, either the geometric series, or an integral.

this one follows immediately from comparison with the integral of x^(-2).

nuff sedd.
 
  • #12
For Abel's test the most obvious breakdown will probably involve showing the sums [itex]\sum_{n=1}^{N}\cos n[/itex] are bounded for all N. This may be difficult so a little hint: Euler's formula will reduce this to bounding a geometric series. (showing absolute convergence with a comparison test is probably more straightforward in this case though, and actually yields a stronger result).
 
  • #13
hellooo... this series has terms bounded in absolute value by 1/n^2, hence converges absolutely by comparison with the integral of 1/x^2.
 
  • #14
Of course, I just wanted to clarify Abel's test a little bit as applying it here might be beyond the scope of your typical first yeat calculus course (not that I know the OP's background).
 

Related to Testing convergence of a series

What is the definition of convergence of a series?

Convergence of a series refers to the behavior of a sequence of partial sums of the series. If the sequence of partial sums approaches a finite value as the number of terms increases, the series is said to converge. If the sequence of partial sums does not approach a finite value, the series is said to diverge.

How do you test for convergence of a series?

There are several tests that can be used to determine the convergence of a series, including the ratio test, the root test, and the comparison test. These tests involve evaluating the behavior of the terms in the series and comparing them to known convergent or divergent series.

What is the difference between absolute and conditional convergence?

Absolute convergence refers to a series where the sum of the absolute values of the terms converges. Conditional convergence refers to a series where the sum of the terms converges, but the sum of the absolute values of the terms does not converge. In other words, the order in which the terms are added affects the convergence of the series.

Can a series converge to more than one value?

No, a series can only converge to one value. If a series has multiple limits, it is said to be oscillating and does not converge.

What happens if a series diverges?

If a series diverges, it means that the sequence of partial sums does not approach a finite value. This could be due to the terms in the series increasing without bound, or alternating in a way that prevents the sequence from approaching a finite value. In either case, the series is considered to have no sum.

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