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

Solution:

Step 1:

To determine the sum of an infinite geometric series, we use the formula $S=\frac{a}{1-r}$, where $S$ is the sum, $a$ is the first term, and $r$ is the common ratio.

Step 2:

Calculate the common ratio $r$ by using the formula $r=\frac{a_{n+1}}{a_n}$.

Step 2.1:

Insert the terms $a_n$ and $a_{n+1}$ into the formula to find $r$. $r=\frac{2^{n+1}}{2^n}$

Step 2.2:

Eliminate the same factors in $2^{n+1}$ and $2^n$.

Step 2.2.1:

Extract $2^n$ from $2^{n+1}$. $r=\frac{2^n\cdot 2}{2^n}$

Step 2.2.2:

Remove the identical factors.

Step 2.2.2.1:

Apply multiplication by $1$. $r=\frac{2^n\cdot 2}{2^n\cdot 1}$

Step 2.2.2.2:

Cross out the common factor. $r=\frac{\cancel{2^n}\cdot 2}{\cancel{2^n}\cdot 1}$

Step 2.2.2.3:

Simplify the expression. $r=\frac{2}{1}$

Step 2.2.2.4:

Divide $2$ by $1$. $r=2$

Step 3:

Determine if the series converges. Since $|r|\ge 1$, the series does not converge.

Knowledge Notes:

Infinite geometric series are sequences of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio ($r$). The sum of an infinite geometric series can be found if $|r|<1$ using the formula $S=\frac{a}{1-r}$, where $a$ is the first term of the series. If $|r|\ge 1$, the series does not have a finite sum and is considered to diverge.

In this problem, the series is $2^n$ for $n$ starting from 1 to infinity. This is a geometric series where each term is twice the previous term, indicating that the common ratio $r$ is 2. Since the common ratio is greater than 1, the series diverges, meaning it does not sum to a finite value.