Test the following series for convergence: (i) ∑(n+1)/(n·3^n) from n=1 to ∞ and (ii) ∑1/(n^(1/2) + 2n^(1/4) + 3) from n=1 to ∞. Also, show that the complement of an open set S in R... Test the following series for convergence: (i) ∑(n+1)/(n·3^n) from n=1 to ∞ and (ii) ∑1/(n^(1/2) + 2n^(1/4) + 3) from n=1 to ∞. Also, show that the complement of an open set S in R is closed, and prove that the function f(x) = x^2 is Riemann integrable. Read more

Steps to Solve

1. Testing the first series for convergence

   We analyze the series: $$ \sum_{n=1}^{\infty} \frac{n+1}{n \cdot 3^n} $$

   We can simplify the term: $$ \frac{n+1}{n \cdot 3^n} = \frac{n}{n \cdot 3^n} + \frac{1}{n \cdot 3^n} = \frac{1}{3^n} + \frac{1}{n \cdot 3^n} $$

   This means we can evaluate the convergence of each part:

   • The first part, $\sum_{n=1}^{\infty} \frac{1}{3^n}$, is a geometric series with $r = \frac{1}{3}$, which converges.
   • For the second part, we can use the comparison test with $\sum_{n=1}^{\infty} \frac{1}{3^n}$, recognizing that $\frac{1}{n \cdot 3^n}$ converges because $\frac{1}{3^n}$ dominates $\frac{1}{n \cdot 3^n}$ for large $n$.

   Therefore, the first series converges.

2. Testing the second series for convergence

   Now we evaluate: $$ \sum_{n=1}^{\infty} \frac{1}{n^{1/2} + 2n^{1/4} + 3} $$

   For large $n$, the term $n^{1/2}$ dominates, so we can approximate: $$ n^{1/2} + 2n^{1/4} + 3 \sim n^{1/2} $$

   Thus, we examine: $$ \sum_{n=1}^{\infty} \frac{1}{n^{1/2}} $$

   This series diverges since it is a p-series with $p = \frac{1}{2} < 1$. Therefore, by the limit comparison test, the original series diverges.

3. Proving the complement of an open set is closed

   Let $S$ be an open subset of $\mathbb{R}$. The complement $S^c$ consists of all points in $\mathbb{R}$ not in $S$.

   To show that $S^c$ is closed, we must show that it contains all its limit points. If a point $x$ is a limit point of $S^c$, then every neighborhood of $x$ contains points not in $S$. Since $S$ is open, there must exist an interval around points in $S$, meaning $x$ cannot be in $S$. Thus, $S^c$ contains all its limit points, showing it is closed.

4. Proving the function is Riemann integrable

   The function $f(x) = x^2$ is continuous on the closed interval $[0, 1]$. A fundamental theorem states that continuous functions on closed intervals are Riemann integrable.

   Therefore, we can compute the integral: $$ \int_0^1 x^2 , dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} $$