Question

For what values of p is the following series convergent?

\(\sum\limits_{n = 1}^\infty {\frac{{{{\left( {{\rm{ - }}1} \right)}^{n{\rm{ - }}1}}}}{{{n^p}}}} \)

Short answer

The values are all values greater than or i.e p >0

Step-by-step solution

Given And To Find

\(\sum\limits_{n = 1}^\infty {\frac{{{{\left( {{\rm{ - 1}}} \right)}^{n{\rm{ - }}1}}}}{{{n^p}}}} \)

(given)

To find the values of p. (to find)

Test converges if  \(\frac{1}{{{n^p}}}\)when p > 0      

When p will be greater than 0, then series become alternating series and terms are in decreasing order as \(\frac{1}{{{n^p}}}{\rm{ > }}\frac{1}{{{{\left( {n + 1} \right)}^p}}}\)so, \(_{{{\lim }_{n \to \infty }}\frac{1}{{^{{n^p}}}} = 0}\)

Therefore, by alternating series the series is convergent.

Test limit of  (-1)n-1, when p=0

When we take p=0 the series become \(\sum\limits_{n = 1}^\infty {\frac{{{{\left( {{\rm{ - }}1} \right)}^{n{\rm{ - }}1}}}}{{{n^0}}} = \sum\limits_{n = 1}^\infty {{{\left( {{\rm{ - }}1} \right)}^{n{\rm{ - }}1}} = {\rm{ - }}1 + 1{\rm{ - }}1 + 1....} } \)

Where ^nthterm will be so \(\mathop {\lim }\limits_{n \to \infty } {a_n} = \mathop {\lim }\limits_{n \to \infty } {\left( {{\rm{ - }}1} \right)^{n - 1}}\) which does not tend to be a unique value, so \(\mathop {\lim }\limits_{n \to \infty } {\left( {{\rm{ - }}1} \right)^{n{\rm{ - }}1}}\)does not exist.

To test the limit when p<0

When we take p less than 0than the lim will be \(\mathop {\lim }\limits_{n \to \infty } \frac{1}{{^{{n^p}}}} = \infty \)so the n^thterm will be \({a_n} = \frac{{{\rm{ - }}1}}{{{n^p}}}\)and when applying limit on an we get \(\mathop {\lim }\limits_{n \to \infty } {a_n} = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {{\rm{ - }}1} \right)}^{n{\rm{ - }}1}}}}{{{n^p}}}\)which again does not exist as it tends to ∞or -∞, so by the divergence test the given series is divergent.

Hence, the given series is convergent only when p is greater than 0 that is p>0.