Key Concepts and Formulas

• Geometric Series: The sum of an infinite geometric series with first term $a$ and common ratio $r$ (where $|r| < 1$) is given by $\frac{a}{1-r}$.
• Logarithmic Series Expansion: For $|x| < 1$, the Maclaurin series for $\ln(1-x)$ is given by: $\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - ...$ This implies that: $-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + ...$
• Properties of Exponents and Logarithms: $e^{a+b} = e^a e^b$, $e^{\ln u} = u$, and $e^{-\ln u} = e^{\ln u^{-1}} = u^{-1}$.

Step-by-Step Solution

1. Rewrite the general term of the series: The given series is $y = \frac{1}{2}x^2 + \frac{2}{3}x^3 + \frac{3}{4}x^4 + ...$. We can observe that the coefficient of $x^n$ is $\frac{n-1}{n}$ for $n \ge 2$. Alternatively, and more usefully for this problem, we can rewrite the coefficient of $x^n$ as $\frac{n-1}{n} = 1 - \frac{1}{n}$. However, the series starts with $x^2$, so the index needs careful consideration. Let's write the general term as $\frac{k}{k+1}x^{k+1}$ where $k$ starts from 1. So, $y = \sum_{k=1}^{\infty} \frac{k}{k+1}x^{k+1}$. We can rewrite the coefficient $\frac{k}{k+1}$ as $1 - \frac{1}{k+1}$. Thus, $y = \sum_{k=1}^{\infty} \left(1 - \frac{1}{k+1}\right)x^{k+1}$. Why this step? This manipulation helps in splitting the series into parts that resemble known series expansions.

2. Separate the series into two parts: Expanding the expression from Step 1, we get: $y = \sum_{k=1}^{\infty} x^{k+1} - \sum_{k=1}^{\infty} \frac{1}{k+1}x^{k+1}$ $y = (x^2 + x^3 + x^4 + ...) - \left(\frac{1}{2}x^2 + \frac{1}{3}x^3 + \frac{1}{4}x^4 + ...\right)$ Why this step? This separation allows us to identify a geometric series and a series that can be related to the logarithmic series.

3. Evaluate the first series (Geometric Series): The first part is $x^2 + x^3 + x^4 + ...$. This is an infinite geometric series with the first term $a = x^2$ and the common ratio $r = x$. Since $0 < x < 1$, the series converges. The sum of this geometric series is $\frac{a}{1-r} = \frac{x^2}{1-x}$. So, $y = \frac{x^2}{1-x} - \left(\frac{1}{2}x^2 + \frac{1}{3}x^3 + \frac{1}{4}x^4 + ...\right)$. Why this step? We have found a closed-form expression for the first part of the series.

4. Relate the second series to the Logarithmic Series: Consider the logarithmic series expansion: $\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - ...$ Multiplying by -1, we get: $-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + ...$ The second part of our series for $y$ is $\frac{1}{2}x^2 + \frac{1}{3}x^3 + \frac{1}{4}x^4 + ...$. We can see that this is almost $-\ln(1-x)$, but it's missing the first term $x$. So, we can write: $\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + ... = \left(x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + ...\right) - x$ $\frac{1}{2}x^2 + \frac{1}{3}x^3 + \frac{1}{4}x^4 + ... = -\ln(1-x) - x$ Why this step? This step cleverly uses the known logarithmic series to express the second part of $y$ in a closed form.

5. Combine the results to find a closed form for y: Substitute the expressions from Step 3 and Step 4 back into the equation for $y$: $y = \frac{x^2}{1-x} - (-\ln(1-x) - x)$ $y = \frac{x^2}{1-x} + \ln(1-x) + x$ To simplify further, we can combine the first two terms: $y = \frac{x^2 + x(1-x)}{1-x} + \ln(1-x)$ $y = \frac{x^2 + x - x^2}{1-x} + \ln(1-x)$ $y = \frac{x}{1-x} + \ln(1-x)$ Why this step? We have successfully expressed $y$ as a function of $x$ in a compact form.

6. Substitute the value of x: We are given $x = \frac{1}{2}$. Substitute this value into the expression for $y$: $y = \frac{\frac{1}{2}}{1 - \frac{1}{2}} + \ln\left(1 - \frac{1}{2}\right)$ $y = \frac{\frac{1}{2}}{\frac{1}{2}} + \ln\left(\frac{1}{2}\right)$ $y = 1 + \ln\left(\frac{1}{2}\right)$ Using the property $\ln(\frac{a}{b}) = \ln a - \ln b$, we get $\ln(\frac{1}{2}) = \ln 1 - \ln 2 = 0 - \ln 2 = -\ln 2$. So, $y = 1 - \ln 2$. Why this step? We are now ready to calculate the final expression we need to evaluate.

7. Calculate $e^{1+y}$: We need to find the value of $e^{1+y}$. Substitute the value of $y$ found in Step 6: $1 + y = 1 + (1 - \ln 2) = 2 - \ln 2$ Now, calculate $e^{1+y}$: $e^{1+y} = e^{2 - \ln 2}$ Using the property $e^{a-b} = \frac{e^a}{e^b}$: $e^{2 - \ln 2} = \frac{e^2}{e^{\ln 2}}$ Since $e^{\ln 2} = 2$: $e^{1+y} = \frac{e^2}{2}$ Why this step? This is the final calculation to arrive at the answer.

Common Mistakes & Tips

• Index Mismanagement: Be careful with the starting index of the series, especially when relating it to standard series expansions like the geometric or logarithmic series. A common error is to misalign the terms.
• Logarithmic Series Form: Remember that the standard $\ln(1-x)$ series starts with $-x$. If your series has terms like $\frac{x^2}{2}, \frac{x^3}{3}, ...$, you will need to add and subtract $x$ to match the $\ln(1-x)$ formula.
• Algebraic Simplification: Ensure careful algebraic manipulation when combining terms and simplifying expressions, particularly with fractions and logarithms.

Summary

The problem requires us to evaluate $e^{1+y}$ where $y$ is defined by an infinite series. The strategy involves rewriting the general term of the series to split it into a geometric series and a series related to the Taylor expansion of $\ln(1-x)$. By finding closed-form expressions for these parts, we obtain a simplified expression for $y$. Substituting the given value of $x = \frac{1}{2}$ into this expression for $y$ allows us to calculate $1+y$, which can then be used to find $e^{1+y}$ using properties of exponents and logarithms.

The final answer is $\boxed{\frac{1}{2}e^2}$.