Infinite Series Practice Problem 2439 

\(\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{1}{k+1}  \frac{1}{k+2} \right] } }\) 
Determine the convergence or divergence of the series. 
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Choosing Which Tests To Try 

Because of the presence of two terms with a negative sign between them, our first thought is that we might have a telescoping series. So we will try this before going on.
Applying The Tests 

This series looks like a telescoping series. To see if that is what we have, we will build a table containing the first few values and see if we can get some cancellation.
k  a_{k} 

1  \(\displaystyle{ \frac{1}{2}  \frac{1}{3} }\) 
2  \(\displaystyle{ \frac{1}{3}  \frac{1}{4} }\) 
3  \(\displaystyle{ \frac{1}{4}  \frac{1}{5} }\) 
. 

n1  \(\displaystyle{ \frac{1}{n}  \frac{1}{n+1} }\) 
n  \(\displaystyle{ \frac{1}{n+1}  \frac{1}{n+2} }\) 
We can see from this table that the first term in each row is canceled by the second term in the previous row. This means that if we add all the terms together from \(k=1\) to \(k=n\) we will have \(1/2\) from the first row and \(1/(n+2)\) in the last row, giving us the partial sum \(\displaystyle{ S_n = \frac{1}{2}  \frac{1}{n+2} }\).
Taking the limit of \(S_n\) will give us the value to which the series converges.
\(\displaystyle{ \lim_{n\to\infty}{S_n} = }\) \(\displaystyle{ \lim_{n\to\infty}{\left[ \frac{1}{2}  \frac{1}{n+2}\right]} = }\)
\(\displaystyle{ \frac{1}{2}  0 = \frac{1}{2} }\)
Therefore the telescoping series converges to \(1/2\).
Test Summary List and Answer 

test/series  works?  notes  



 


 


 


 

yes 
best test  


 


 


 


 



Final Answer 

The telescoping series \(\displaystyle{ \sum_{k=1}^{\infty}{ \left[ \frac{1}{k+1}  \frac{1}{k+2} \right] } }\) converges to \(1/2\).
Notes 

Due to the nature of the \(a_n\) terms, it was not necessary to try any other test. We may have been able to use other tests to determine convergence but they would not have given us the value to which the series converges.
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