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In Exercises 113 and \(114,\) find the values of \(p\) for which the series converges. $$ \sum_{n=1}^{\infty}(-1)^{n}\left(\frac{1}{n+p}\right) $$

Short Answer

Expert verified
The series converges for all values of \(p\) where \(p > -1\).

Step by step solution

01

Identify the Series Type

Identify the series as an alternating series. It alternates signs because of the \(-1^n\) term and it has a general term \(\frac{1}{n+p}\).
02

Identify the Decaying Function

The decaying function is \(\frac{1}{n+p}\). Compare this to the general term of an alternating series, \(b_n\).
03

Apply the First Condition of the Alternating Series Test

The first condition requires that the series must be decreasing. Check to see if \(b_{n+1} \leq b_n\) for all \(n\) when \(p > -1\). This is because the denominator \(n+p\) needs to be positive.
04

Apply the Second Condition of the Alternating Series Test

The second condition requires that \(\lim_{n \rightarrow \infty}b_n = 0\). Calculate the limit and verify that it's equal to zero when \(p > -1\).
05

Conclusion

The two conditions of the Alternating Series Test are met if \(p > -1\), so the series converges for \(p > -1\). If \(p \leq -1\), either the series is not decreasing or the limit is not zero.

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