Understanding the nth-term test for Divergence

In summary: Divergence.The Nth Term Test for Divergence is a test that applies to a series only if the series does not converge to 0.
  • #1
chwala
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TL;DR Summary
I am looking at the theorem;

It states that 'If the sequence ##a_n## does not converge to 0, then the series ##\sum_{n=0}^\infty a_n## diverges.
There are several examples that i have looked at which are quite clear and straightforward, e.g

##\sum_{n=0}^\infty 2^n##

it follows that

##\lim_{n \rightarrow \infty} {2^n}=∞##

thus going with the theorem, the series diverges.

Now let's look at the example below;

##\sum_{n=1}^\infty \dfrac{1}{n}##

it follows that,

##\lim_{n \rightarrow \infty} {\left[\dfrac{1}{n}\right]}=0##

the text indicates that the nth term test for divergence does not apply...infact we know that the series diverges (using p -series)...

now back to my question.

What is the relevance of this theorem, if any ...why use 'does not converge to 0, if converging to 0 after all has no implication' whatsoever? the theorem is not complete by itself in my thinking.

Can we say converging to 0, may or may not mean convergence? implying that we have to use other tools to ascertain that?

I am also looking also at the nth-term test of a convergent series- Now if a series ##b_n## diverges, does the sequence ##b_n## converge to 0 or not? just going with what is stated in the theorems.

thanks.
 
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  • #2
Looking at the limit of a sequence is ”easier” than looking at the limit convergence of a series. This test allows you to trivially answer a lot of series. You then need other approaches for the remainder.

1/n is also a counterexample for your “series then sequence“ question.
 
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  • #3
Frabjous said:
Looking at the limit of a sequence is ”easier” than looking at the limit of series. This test allows you to trivially answer a lot of series.

1/n is also a counterexample for your “series then sequence“ question.
What do you mean by 'limit of a series' ?
 
  • #4
chwala said:
What do you mean by limit of a series?
Convergence
 
  • #5
Frabjous said:
Looking at the limit of a sequence is ”easier” than looking at the limit of series. This test allows you to trivially answer a lot of series. You then need other approaches for the remainder.

1/n is also a counterexample for your “series then sequence“ question.
How would you to determine the convergence or divergence of ##\sum_{n=1}^\infty \dfrac{1}{n}## by looking at limits of its sequence?
 
  • #6
chwala said:
How would you use only limits of a given sequence as you indicate to determine convergence or divergence of a given series?

you have a different explanation or approach?
There are also the ratio and root tests, but they are inconclusive if the ratio/root is 1.
 
  • #7
Frabjous said:
There are also the ratio and root tests, but they are inconclusive if the ratio/root is 1.
I am aware and also conversant with the other tests for determining convergence or divergence of both series and sequences...my interest is solely on the theorem as it is...

the nth terms test for both convergence and for divergence are is restricted to only give a particular direction...and the contra statement (opposite) may not apply...that is the query that i had. Cheers and thanks mate.
 
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  • #8
chwala said:
'If the sequence ##a_n## does not converge to 0, then the series ##\sum_{n=0}^\infty a_n## diverges.
Maybe the confusion is this...

The theorem applies only when ##a_n## does not converge to zero.

A ‘more complete’ statement of the theorem might be, for example:

If the sequence ##a_n## does not converge to 0, then the series ##\sum_{n=0}^\infty a_n## diverges.​
If the sequence ##a_n## does converge to 0, then the series ##\sum_{n=0}^\infty a_n## may converge or diverge.​
 
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  • #9
Steve4Physics said:
Maybe the confusion is this...

The theorem applies only when ##a_n## does not converge to zero.

A ‘more complete’ statement of the theorem might be, for example:

If the sequence ##a_n## does not converge to 0, then the series ##\sum_{n=0}^\infty a_n## diverges.​

If the sequence ##a_n## does converge to 0, then the series ##\sum_{n=0}^\infty a_n## may converge or diverge.​
The theorem as it is 'is confusing'.

Yap- this is now clear to me...i will add it to my notes. Thanks @Steve4Physics
 
  • #10
chwala said:
the nth terms tests for both convergence and divergence are restricted to only give a particular direction
Textbooks often call this theorem the Nth Term Test for Divergence. The test applies only to series for which ##\lim_{n \to \infty}a_n \ne 0##. There is no nth term test for convergence.
 
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  • #11
Hi @Mark44 , I think we do have the nth term theorem for convergence.
 
  • #12
chwala said:
I think we do have the nth term theorem for convergence.
What is it?
 
  • #13
Theorem: Limit of nth term of a Convergent Series

If the series ##\sum{a_n}## converges, then the sequence ##a_n## converges to 0....the contrapositive being the nth-term test for Divergence.
 
  • #14
chwala said:
Theorem: Limit of nth term of a Convergent Series

If the series ##\sum{a_n}## converges, then the sequence ##a_n## converges to 0.
A better name would be the Convergent Series test.
chwala said:
the contrapositive being the nth-term test for Divergence.
Yes, but the Nth term test for divergence is usually phrased in terms of the limit of the nth term of some series.
 
  • #15
I like this proof for ##\Sigma \frac{1}{n} \rightarrow \infty##, which may be applicable to proving other convergence/divergence too:

$$\Sigma \frac{1}{n}$$

$$ = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15} + \frac{1}{16} + ...$$

$$= (\frac{1}{2}) + (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}) + (\frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15} + \frac{1}{16}) + ...$$

$$ > (\frac{1}{2}) + (\frac{1}{4} + \frac{1}{4}) + (\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}) + (\frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16}) + ...$$

$$ = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ...$$

$$ \rightarrow \infty$$
 
  • #16
James1238765 said:
I like this proof for ##\Sigma \frac{1}{n} \rightarrow \infty##, which may be applicable to proving other convergence/divergence too:

$$\Sigma \frac{1}{n}$$

$$ = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15} + \frac{1}{16} + ...$$

$$= (\frac{1}{2}) + (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}) + (\frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15} + \frac{1}{16}) + ...$$

$$ > (\frac{1}{2}) + (\frac{1}{4} + \frac{1}{4}) + (\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}) + (\frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16}) + ...$$

$$ = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ...$$

$$ \rightarrow \infty$$
Nice one...understanding of p- series is of essence on this kind of problems though...
 

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