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17Calculus Infinite Series - Practice Problem 2439

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.
If the series converges,
- determine the value the series converges to, if possible; and
- determine if the series converges absolutely or conditionally.

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

Telescoping Series

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.




\(\displaystyle{ \frac{1}{2} - \frac{1}{3} }\)


\(\displaystyle{ \frac{1}{3} - \frac{1}{4} }\)


\(\displaystyle{ \frac{1}{4} - \frac{1}{5} }\)



\(\displaystyle{ \frac{1}{n} - \frac{1}{n+1} }\)


\(\displaystyle{ \frac{1}{n+1} - \frac{1}{n+2} }\)

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




divergence test


geometric series

alternating series test

telescoping series


best test

ratio test

limit comparison test

direct comparison test

integral test

root test

Final Answer

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


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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