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Determine whether the following series is an absolute convergence, conditional convergence, or divergent and by what test:

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2 Answers

Parentheses Matter!

Your description of the problem to has unbalanced parentheses, so there is ambiguity regarding exactly what series the question is about. The + 1 at the end could be:
  1. completely outside the summation:  1 + ∑n=1 ((-1)n(n+1)2/n5) ;
  2. inside the summation, as a separate term: ∑n=1 (1 + (-1)n(n+1)2/n5) ; or
  3. inside the summation, as part of the denominator:  ∑n=1 ((-1)n(n+1)2/(n5 + 1)) ;
The third option is the most likely, only because the other two options are less interesting.
 
Since n is strictly positive, the summand has the form (-1)nbn where the bn ≥ 0. Thus, we could show that this series converges using the Alternating Series Test, which merely requires that we show that:
  1. limnbn = 0; and
  2. {bn} is a decreasing sequence.
However, the query wants to know whether the series is Absolutely Convergent. Thus, to determine this we must examine the series consisting of the absolute values of the original terms, which in this case is just  ∑n=1bn If this series converges, then the original series is converges absolutely.
 
An easy way to establish the convergence of this series is to identify another series whose terms are all larger the terms in our series, that we know converges. The second series establishes an upper bound on the value of our series, and so its convergence guarantees the convergence of our series. This is called the Comparison Test.
 
Note the following:
  1. for n>2, (n+1)2n3 ; and 
  2. (n5+1) > n5
therefore, 
 
(n + 1)2/(n5 + 1) < n3/n51/n2
 
Thus, our series is absolutely convergent if ∑n=1 (1/n2) converges.
 
But, this is a series that is known to converge. In fact, we know that its value is π2/6 (pi squared divided by 6).
The convergence of this series can be shown using the Integral Test, or more directly, by the p-series Test,
since it has the form:
 
n=k (1/np) for k > 0 ( = 1 in our case) and p > 1 (= 2 in our case).
 
Thus, our original series is absolutely convergent.
 
 
 
 
I'm assuming the +1 is still in the denominator. In this case, we can compare the series with one we know converges:
Since
(n+1)2<n3,  n=3,4,5,... (doesn't matter that the inequality doesn't hold for n=1,2)
and
n5+1>n5, so 1/(n5+1)<1/n5, n=1,2,3,...
we know
(n+1)2/(n5+1) < n3/n5=1/n2,  n=3,4,5,...
 
By the p-series test, we know that ∑n=1 1/n² converges. Therefore, by the comparison test,
n=1 (n+1)2/(n5+1) converges.
Therefore,
n=1 (-1)n(n+1)2/(n5+1) converges absolutely.