The mathematical expression given in the image is:

$\sum_{n=1}^{\infty} \frac{2n}{3^n}$

This is an infinite series, and to solve it, we can break it down step by step.

General Approach:

We are summing terms of the form $\frac{2n}{3^n}$, where $n$ ranges from 1 to infinity.

This series can be handled using a standard summation technique for series of the form $\sum_{n=1}^{\infty} n x^n$, where $x$ is a constant ratio. The formula for this kind of series is:

$\sum_{n=1}^{\infty} n x^n = \frac{x}{(1 - x)^2} \quad \text{for} \quad |x| < 1$

In our case, we have $x = \frac{1}{3}$, and we can adjust the formula to account for the factor $2$.

Let me calculate the result step by step.The value of the infinite series

$\sum_{n=1}^{\infty} \frac{2n}{3^n}$

is $\frac{3}{2}$.

Would you like more details on how this result was derived, or do you have any other questions?

Here are 5 related questions to explore further:

1. How would the result change if the denominator were $4^n$ instead of $3^n$?
2. What is the general formula for sums involving $n$ in the numerator?
3. Can this type of sum be applied to geometric progressions with different starting indices?
4. What happens if we add a constant term to the summand, such as $\frac{2n + 1}{3^n}$?
5. How can you determine the convergence of an infinite series in general?

Tip: When dealing with infinite series, always check the convergence criteria, especially if the series involves ratios or powers.