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.