Question

How do you determine if the series the converges conditionally, absolutely or diverges given
from $[1,\infty )$?

Answer 1

Hint: In this question we will simplify the expression based on the value of what $\cos (n\pi )$ will give us based on the value of $n$ and then we will simplify the expression into the form of a series check whether the series is decreasing or increasing, based on that we will give the final conclusion whether the series is converges conditionally, absolutely or diverges in the series.

Complete step-by-step answer:
We have the given series as $\sum{\dfrac{\cos (n\pi )}{n+1}}$which has the range of $[1,\infty )$
Therefore, the series can be written in the summation form as:
$\Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{\cos (n\pi )}{n+1}}$
Now we know that $\cos (\pi )=-1$ and that of $\cos (2\pi )=1$ and this pattern keeps repeating for every increment in the value of $n$. This means that the value is $-1$ whenever the value of $n$ is odd and has a value $1$ whenever the value of $n$ is even. Therefore, we can generalize the solution $\cos (n\pi )={{(-1)}^{n}}$ since a negative number raised to an even number will yield a positive number.
On substituting the value in the expression, we get:
$\Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{{{(-1)}^{n}}}{n+1}}$
Now we will check whether the series is absolutely convergent, for a series to be absolutely convergent the summation of the absolute value of all the summands is a finite term, therefore:
$\Rightarrow \sum\limits_{n=1}^{\infty }{\left| \dfrac{{{(-1)}^{n}}}{n+1} \right|}$
on expanding the series with only the positive terms, we get:
$\Rightarrow \sum\limits_{n=1}^{\infty }{\left| \dfrac{{{(-1)}^{n}}}{n+1} \right|}=\sum\limits_{n=1}^{\infty }{\dfrac{1}{n+1}=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}.....}$
Which is the harmonic series, we know that the harmonic series is divergent therefore the expression cannot be absolutely convergent.
Now we will check whether the expression is conditionally convergent. For a series to be conditionally convergent the sum of the positive terms diverges to positive infinity or sum of negative terms lead to negative infinity.
Now we can see the sum of positive terms as:
$\Rightarrow \sum\limits_{n=1}^{\infty }{\dfrac{1}{n+1}}$.
Since $\dfrac{1}{n+1}$ is decreasing, $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n+1}=0$ therefore, we know the series is conditionally convergent, which is the required solution.

Note: It is to be remembered that the harmonic series is a divergent series. The harmonic series is represented as: $\sum\limits_{n=1}^{\infty }{\dfrac{1}{n}}$. The common mistake done in questions of series is that the denominator of the fraction should never be zero because it will result in a fallacy.
The basic type of series in mathematics are arithmetic series in which the difference between two consecutive terms is a constant and geometric series in which the ratio between two consecutive terms is a constant.