Find Limit of 1/2 Series: 1/(n^2 + n)

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SUMMARY

The limit of the series 1/(n^2 + n) from n=1 to infinity converges to 1/2. This conclusion is derived from recognizing the series as a telescoping series, where the terms can be simplified to reveal the limit. The area of triangles formed between 1/n and 1/(n+1) contributes to this limit, confirming that the integral of the oscillating function aligns with the series' behavior. The analysis confirms that the limit is straightforward to determine through proper simplification techniques.

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Homework Statement



What is the limit of 1/2 Series from (n=1 to n=Infinity) of 1/(n^2 + n).


The Attempt at a Solution



This is a simplification of finding the integral of an oscillating function from 0 to 1 that makes triangles of height one between 1/n and 1/(n+1)

Thus the area of each triangle is 1/2 * (1/n - 1/(n+1)) = 1/(2n^2 + 2n)

The integral should therefore be equal to the above given limit. From intuition I believe the answer should be 1/2, though I would appreciate any help in determining how this limit is found. Seems like it should be relativley easy to find.
 
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Well, it's easier in the original form. You want to sum 1/n - 1/(n+1) from n=1 to infinity. Let's write out the first few terms:
[tex]\left({1\over 1}-{1\over 2}\right)<br /> +\left({1\over 2}-{1\over 3}\right)<br /> +\left({1\over 3}-{1\over 4}\right)+\ldots[/tex]
Notice any way to simplify this?
 
As Avodyne said, it's a telescoping series
 

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