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 Jan 21, 2017

Best Answer 

 #1
avatar+128408 
+5

1. 

 

inf

∑   2 / [ n ( n + 1) ]     converges to  2   

1

 

This series converges by a comparison test  to  2 / n^2 [ which conveges by the integral test ]

 

2 / n^2                       2       1/2     2/9  ..........      

2 / [ n ( n + 1) ]          1       1/3     1/6  ..........

 

Both series  >  0  for all terms....and, term by term, the second series is smaller than the second

 

So....if 2/n^2 converges....so does  2 / [ n ( n + 1) ] 

 

 

2.

 

inf

∑   [ n / ( n^2 -  1) ]    

2

 

This series diverges by a comparison test  to 1 / n  [ which diverges by the p - series test]

 

1/n                         1/2        1/3      1/4  .........

n / ( n^2 -  1)         2/3        3/8       4/15  ........

 

Both series  >  0  for all terms....and, term by term, the first series is smaller than the second

 

So....if 1/n diverges....so does  n / ( n^2 -  1)  

 

 

 

cool cool cool

 Jan 21, 2017
 #1
avatar+128408 
+5
Best Answer

1. 

 

inf

∑   2 / [ n ( n + 1) ]     converges to  2   

1

 

This series converges by a comparison test  to  2 / n^2 [ which conveges by the integral test ]

 

2 / n^2                       2       1/2     2/9  ..........      

2 / [ n ( n + 1) ]          1       1/3     1/6  ..........

 

Both series  >  0  for all terms....and, term by term, the second series is smaller than the second

 

So....if 2/n^2 converges....so does  2 / [ n ( n + 1) ] 

 

 

2.

 

inf

∑   [ n / ( n^2 -  1) ]    

2

 

This series diverges by a comparison test  to 1 / n  [ which diverges by the p - series test]

 

1/n                         1/2        1/3      1/4  .........

n / ( n^2 -  1)         2/3        3/8       4/15  ........

 

Both series  >  0  for all terms....and, term by term, the first series is smaller than the second

 

So....if 1/n diverges....so does  n / ( n^2 -  1)  

 

 

 

cool cool cool

CPhill Jan 21, 2017
 #2
avatar+97 
+5

Thanks CPhill!!laugh

Voncave  Jan 21, 2017

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