a. Find the equation of the tangent plane to $z = \ln(2x+y)$ at the point $(1, 2)$. b. Let $\sum_{x=1}^{\infty} \frac{1}{x^p}$ be a $p$-series. Show that the series converges if $p > 1$ and diverges if $0 < p \leq 1$.

1. Problem Description

a. Find the equation of the tangent plane to $z = \ln(2x+y)$ at the point $(1, 2)$.

b. Let $\sum_{x=1}^{\infty} \frac{1}{x^p}$ be a $p$-series. Show that the series converges if $p > 1$ and diverges if $0 < p \leq 1$.

2. Solution Steps

a. To find the equation of the tangent plane to the surface $z = f(x, y)$ at the point $(x_0, y_0)$, we use the formula:

$z - f(x_0, y_0) = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$

First, we find $f(1, 2)$:

$f(1, 2) = \ln(2(1) + 2) = \ln(4)$

Next, we find the partial derivatives $f_x$ and $f_y$:

$f_x(x, y) = \frac{\partial}{\partial x} \ln(2x + y) = \frac{2}{2x + y}$

$f_y(x, y) = \frac{\partial}{\partial y} \ln(2x + y) = \frac{1}{2x + y}$

Now, we evaluate the partial derivatives at $(1, 2)$:

$f_x(1, 2) = \frac{2}{2(1) + 2} = \frac{2}{4} = \frac{1}{2}$

$f_y(1, 2) = \frac{1}{2(1) + 2} = \frac{1}{4}$

Plugging these values into the tangent plane equation:

$z - \ln(4) = \frac{1}{2}(x - 1) + \frac{1}{4}(y - 2)$

$z = \frac{1}{2}x - \frac{1}{2} + \frac{1}{4}y - \frac{1}{2} + \ln(4)$

$z = \frac{1}{2}x + \frac{1}{4}y - 1 + \ln(4)$

b. Consider the $p$-series $\sum_{n=1}^{\infty} \frac{1}{n^p}$.

Case 1: $p > 1$. We can use the integral test. Consider the integral $\int_{1}^{\infty} \frac{1}{x^p} dx$.

$\int_{1}^{\infty} \frac{1}{x^p} dx = \lim_{t \to \infty} \int_{1}^{t} x^{-p} dx = \lim_{t \to \infty} \left[ \frac{x^{-p+1}}{-p+1} \right]_{1}^{t} = \lim_{t \to \infty} \frac{1}{1-p} (t^{1-p} - 1)$.

Since $p > 1$, $1 - p < 0$, so $\lim_{t \to \infty} t^{1-p} = 0$. Therefore, the integral converges to $\frac{1}{p-1}$. By the integral test, the series converges for $p > 1$.

Case 2: $0 < p \leq 1$. Again, we can use the integral test.

$\int_{1}^{\infty} \frac{1}{x^p} dx = \lim_{t \to \infty} \int_{1}^{t} x^{-p} dx = \lim_{t \to \infty} \left[ \frac{x^{-p+1}}{-p+1} \right]_{1}^{t} = \lim_{t \to \infty} \frac{1}{1-p} (t^{1-p} - 1)$.

If $0 < p < 1$, then $1 - p > 0$, so $\lim_{t \to \infty} t^{1-p} = \infty$. The integral diverges.

If $p = 1$, the series is $\sum_{n=1}^{\infty} \frac{1}{n}$, the harmonic series, which is known to diverge. Alternatively, we can use the integral test: $\int_{1}^{\infty} \frac{1}{x} dx = \lim_{t \to \infty} [\ln(x)]_{1}^{t} = \lim_{t \to \infty} (\ln(t) - \ln(1)) = \infty$. The integral diverges.

By the integral test, the series diverges for $0 < p \leq 1$.

3. Final Answer

a. $z = \frac{1}{2}x + \frac{1}{4}y - 1 + \ln(4)$

b. The $p$-series $\sum_{x=1}^{\infty} \frac{1}{x^p}$ converges if $p > 1$ and diverges if $0 < p \leq 1$.