Evaluate the Summation sum from n=1 to infinity of 2n-3

Solution:

Step 1:

First, we need to analyze the given series to determine if it is a convergent or divergent series. The series in question is $\sum _{n=1}^{\mathrm{\infty }}(2n-3)$.

Step 2:

We can rewrite the series as $\sum _{n=1}^{\mathrm{\infty }}2n-\sum _{n=1}^{\mathrm{\infty }}3$. This separates the series into two parts: one that is dependent on $n$ and one that is a constant series.

Step 3:

The first part, $\sum _{n=1}^{\mathrm{\infty }}2n$, is a divergent series because it is a multiple of the harmonic series, which is known to diverge.

Step 4:

The second part, $\sum _{n=1}^{\mathrm{\infty }}3$, is also divergent because it is an infinite sum of a non-zero constant.

Step 5:

Since both parts of the series are divergent, the entire series $\sum _{n=1}^{\mathrm{\infty }}(2n-3)$ is divergent.

Step 6:

Therefore, we conclude that the summation does not converge to a finite value.

Solution:"The series $\sum _{n=1}^{\mathrm{\infty }}(2n-3)$ is divergent."

Knowledge Notes:

1. A series is convergent if the sum of its terms approaches a finite limit as $n$ approaches infinity.

2. A series is divergent if the sum of its terms does not approach a finite limit as $n$ approaches infinity.

3. The harmonic series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}$ is a well-known example of a divergent series.

4. Any series that is a non-zero constant times the harmonic series will also diverge.

5. An infinite sum of a non-zero constant is always divergent because it will grow without bound as more terms are added.

6. In this case, since both parts of the series are divergent, the entire series is divergent.