Question

How do you test the alternating series $ \sum {{{\left( { - 1} \right)}^{n + 1}}\dfrac{n}{{10n + 5}}} $ from $ n $ is $ \left[ {1,\infty } \right) $ for convergence?

Answer 1

Hint: In order to test the above given summation series , use the alternating series test for form $ \sum {{{\left( { - 1} \right)}^n}{b_n}} $ where $ {b_n} $ is a sequence containing positive terms the conditions for convergence is: if
1. $ {b_n} > {b_{n + 1}} $ which means the sequence is ultimately decreases for all value of $ n $
2. $ \mathop {\lim }\limits_{n \to \infty } \,{b_n} = 0 $

Complete step-by-step answer:
We are given a summation series of the form $ \sum {{{\left( { - 1} \right)}^{n + 1}}\dfrac{n}{{10n + 5}}} $ from $ n $ is $ \left[ {1,\infty } \right) $ .
We can rewrite the summation by including the limits in the summation symbol as
 $ \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n + 1}}\dfrac{n}{{10n + 5}}} $
According to the alternating Series test , if we have a series of the form $ \sum {{{\left( { - 1} \right)}^n}{b_n}} $ , where $ {b_n} $ is a sequence which is having positive terms and the series converges if
1. $ {b_n} > {b_{n + 1}} $ which means the sequence is ultimately decreases for all value of $ n $
2. $ \mathop {\lim }\limits_{n \to \infty } \,{b_n} = 0 $
Note that here we don’t need to have the part $ {\left( { - 1} \right)^{n + 1}} $ any term that causes alternating signs, such as $ \cos \left( {n\pi } \right) $ , $ {\left( { - 1} \right)^{n - 1}} $ , $ {\left( { - 1} \right)^{n + 1}} $ , is okay.
Now , here in this case we have $ {b_n} = \dfrac{n}{{10n + 5}} $ . If we take the limit , we have
\[\mathop {\lim }\limits_{n \to \infty } \dfrac{n}{{10n + 5}} = \dfrac{1}{{10}} \ne 0\].Hence the alternating series test is inconclusive her .
Now if we take the limit on overall sequence, we get the result
  $ \mathop {\lim }\limits_{n \to \infty } = {\left( { - 1} \right)^{n + 1}}\dfrac{n}{{10n + 5}} \ne 0 $
From the above result obtained , we can conclude that the limit does not truly exist.
We can conclude from all the results above that the alternate signs make the sequence closer to $ \dfrac{1}{{10}} $ , causing divergence by the divergence test.
Therefore, the given summation series is divergent in nature.

Note: 1. Students must know the meaning of convergence and divergence of a series before proceeding with the solution.
2.The nth partial sum of the series for $ n $ as $ \left[ {1,\infty } \right) $ \[\sum\limits_{n = 1}^\infty {{a_n}} \] is given by $ {S_n} = {a_1} + {a_2} + .... + {a_n} $ .If the sequence of these partial sums $ \left\{ {{S_n}} \right\} $ converges to L, then the sum of the series also converges to L. Similarly, if $ \left\{ {{S_n}} \right\} $ diverges to L , then the sum of the series also diverges.