Convergence of series and limits

  • Thread starter Juggler123
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  • #1
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For the following series sum(an) between n=2, n=infinity determine if the series converges and if so find the limit? Such that

an=1/(n(n^2-1))

I've expressed (an) in partial fractions so that

an=-1/n + 1/(2(n+1)) + 1/(2(n-1))

but can't seem to get any further than this?
Where should I go from here??
 

Answers and Replies

  • #2
Hi Juggler123.

Have you learned the Limit Comparison Test? This is the key to solving for convergence/divergence of this series.

Usually, we use partial fractions when we have reason to believe that the series will become a "telescoping series" (as you write the sequence of partial sums.... you find "patterns" and these "patterns" cancel out).

Like I said, have a try with the Limit Comparison test... comparing the series you've stated with 1 / n^3
 
  • #3
HallsofIvy
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For the following series sum(an) between n=2, n=infinity determine if the series converges and if so find the limit? Such that

an=1/(n(n^2-1))

I've expressed (an) in partial fractions so that

an=-1/n + 1/(2(n+1)) + 1/(2(n-1))

but can't seem to get any further than this?
Where should I go from here??
Looks to me like you might have a "telescoping series" there. Write out the first few terms using your partial fractions:
[itex]a_2+ a_ 3+ a_4+ ...]/itex]
1/2+ 1/6+ 1/12+ ...
(-1/2+ 1/6+ 1/2)+ (-1/3+ 1/8+ 1/4)+ (-1/4+ 1/10+ 1/6)

That's not terribly clear but in each term we have one negative number, -1/n, and two negative numbers, 1/(2(n+1)) and 1/(2(n-1)). Each -1/n will cancel both 1/(2(m+1)), for m= n-1, and 1/(2(m-1)) for m= n+1. For example, taking n= 3, -1/3 will cancel both 1/(2(2+1)= 1/6, and 1/(2(4-1))= 1/6: -1/3+ 1/6+ 1/6= 0. If n= 6, -1/5 will cancel both 1/(2(4+1))= 1/10 and 1/(2(6-1))= 1/10: -1/5+ 1/10+ 1/10= 0. The only question remaining is what numbers do NOT get canceled?
 

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