$$ \text{Evaluate } \lim_{n \to \infty} \frac{5^n n!}{(2n)^n} \text{ using the ratio test for } a_n = \frac{5^n n!}{(2n)^n} \\\begin{aligned}& \text{Compute } \frac{a_{n+1}}{a_n} = \frac{\frac{5^{n+1} (n+1)!}{(2(n+1))^n}}{\frac{5^n n!}{(2n)^n}} = \frac{5^{n+1} (n+1)!}{(2(n+1))^n} \cdot \frac{(2n)^n}{5^n n!} \\& \text{Numerator: } 5^{n+1} (n+1)! = 5 \cdot 5^n \cdot (n+1) \cdot n! \\& \text{Denominator: } (2(n+1))^n = 2^n (n+1)^n \\& \text{Ratio: } \frac{5 \cdot 5^n \cdot (n+1) \cdot n! \cdot (2n)^n}{5^n \cdot n! \cdot 2^n (n+1)^n} \\& \text{Cancel } 5^n, n!\text{: } \frac{5 (n+1) (2n)^n}{2^n (n+1)^n} = 5 (n+1) \cdot \left( \frac{2n}{2(n+1)} \right)^n \\& = 5 (n+1) \cdot \left( \frac{n}{n+1} \right)^n \\& \text{Limit: } \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \left[ 5 (n+1) \cdot \left( \frac{n}{n+1} \right)^n \right] \\& \text{Evaluate } \left( \frac{n}{n+1} \right)^n = \frac{1}{\left(1 + \frac{1}{n}\right)^n}, \ \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e \\& \text{So, } \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^n = \frac{1}{e} \\& \text{Thus, } \lim_{n \to \infty} \left[ 5 (n+1) \cdot \frac{1}{e} \right] = \frac{5}{e} \cdot \lim_{n \to \infty} (n+1) = \infty \\& \text{Since } \frac{a_{n+1}}{a_n} \to \infty, \ a_n \text{ grows without bound.} \\& \text{Therefore, } \lim_{n \to \infty} \frac{5^n n!}{(2n)^n} = \infty \\& \text{Final Answer: } \infty\end{aligned} $$

$$ \text{Evaluate } \lim_{n \to \infty} \frac{n^n}{(n!)^2} \text{ using the ratio test for } a_n = \frac{n^n}{(n!)^2} \\\begin{aligned}& \text{Compute } \frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)^{n+1}}{((n+1)!)^2}}{\frac{n^n}{(n!)^2}} = \frac{(n+1)^{n+1}}{((n+1)!)^2} \cdot \frac{(n!)^2}{n^n} \\& \text{Numerator: } (n+1)^{n+1} \cdot (n!)^2 \\& \text{Denominator: } ((n+1)!)^2 = (n+1)^2 \cdot (n!)^2 \\& \text{Ratio: } \frac{(n+1)^{n+1} \cdot (n!)^2}{(n+1)^2 \cdot (n!)^2 \cdot n^n} = \frac{(n+1)^{n+1}}{(n+1)^2 \cdot n^n} \\& = (n+1)^{n-1} \cdot \frac{1}{n^n} = \frac{1}{n} \cdot \left( \frac{n+1}{n} \right)^{n-1} = \frac{1}{n} \cdot \left( 1 + \frac{1}{n} \right)^{n-1} \\& \text{Limit: } \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \left[ \frac{1}{n} \cdot \left( 1 + \frac{1}{n} \right)^{n-1} \right] \\& \text{Evaluate: } \left( 1 + \frac{1}{n} \right)^{n-1} = \frac{\left( 1 + \frac{1}{n} \right)^n}{\left( 1 + \frac{1}{n} \right)}, \ \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n = e \\& \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^{n-1} = \frac{e}{1} = e \\& \text{Thus: } \lim_{n \to \infty} \left[ \frac{1}{n} \cdot e \right] = 0 \cdot e = 0 \\& \text{Since } \frac{a_{n+1}}{a_n} \to 0 < 1, \ a_n \to 0. \\& \text{Therefore: } \lim_{n \to \infty} \frac{n^n}{(n!)^2} = 0 \\& \text{Final Answer: } 0\end{aligned} $$

$$ \text{Evaluate } \lim_{n \to \infty} \sum_{k=1}^n \frac{k}{3n^2 + k} \text{ using Riemann sum approach} \\\begin{aligned}& \text{Rewrite term: } \frac{k}{3n^2 + k} = \frac{k}{n^2 \left( 3 + \frac{k}{n^2} \right)} = \frac{1}{n} \cdot \frac{\frac{k}{n}}{3 + \frac{k}{n^2}} \\& \text{Let } x_k = \frac{k}{n}, \ \text{so } \frac{k}{n^2} = \frac{x_k}{n}, \ \text{term becomes: } \frac{1}{n} \cdot \frac{x_k}{3 + \frac{x_k}{n}} \\& \text{Sum: } \sum_{k=1}^n \frac{1}{n} \cdot \frac{\frac{k}{n}}{3 + \frac{k}{n^2}} = \sum_{k=1}^n \frac{1}{n} \cdot \frac{x_k}{3 + \frac{x_k}{n}} \\& \text{As } n \to \infty, \ \frac{1}{n} \text{ acts as } \Delta x, \ x_k \text{ from } \frac{1}{n} \text{ to } 1, \ \text{approximating } \int_0^1 f(x) \, dx \\& \text{Where } f_n(x_k) = \frac{x_k}{3 + \frac{x_k}{n}} \to f(x) = \frac{x}{3} \text{ as } \frac{x_k}{n} \to 0 \\& \text{Thus: } \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n} \cdot \frac{\frac{k}{n}}{3 + \frac{k}{n^2}} = \int_0^1 \frac{x}{3} \, dx \\& \text{Compute integral: } \int_0^1 \frac{x}{3} \, dx = \frac{1}{3} \int_0^1 x \, dx = \frac{1}{3} \cdot \left[ \frac{x^2}{2} \right]_0^1 \\& = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} \\& \text{Note: } \frac{x_k}{n} \leq \frac{1}{n} \to 0, \ \text{ensuring uniform convergence.} \\& \text{Final Answer: } \frac{1}{6}\end{aligned} $$

$$ \begin{aligned}&\text{Evaluate } \lim_{n \to \infty} \sum_{k=1}^n \frac{k}{3n^2 + k} \text{ by approximating the denominator.} \\&\text{The given sum is:} \\&\sum_{k=1}^n \frac{k}{3n^2 + k} \\&\text{Approximate } 3n^2 + k \approx 3n^2 \text{ for large } n, \text{ since } k \leq n: \\&\frac{k}{3n^2 + k} \approx \frac{k}{3n^2} \\&\text{Thus:} \\&\sum_{k=1}^n \frac{k}{3n^2 + k} \approx \sum_{k=1}^n \frac{k}{3n^2} = \frac{1}{3n^2} \sum_{k=1}^n k \\&\text{Sum of first } n \text{ integers:} \sum_{k=1}^n k = \frac{n(n+1)}{2} \\&\text{So:} \\&\frac{1}{3n^2} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{6n^2} = \frac{n^2 + n}{6n^2} = \frac{1}{6} + \frac{1}{6n} \\&\text{Limit:} \\&\lim_{n \to \infty} \left( \frac{1}{6} + \frac{1}{6n} \right) = \frac{1}{6} \\&\text{To justify, bound the term since } 3n^2 \leq 3n^2 + k \leq 3n^2 + n: \\&\frac{k}{3n^2 + n} \leq \frac{k}{3n^2 + k} \leq \frac{k}{3n^2} \\&\text{Upper bound:} \\&\sum_{k=1}^n \frac{k}{3n^2} = \frac{n(n+1)}{6n^2} \to \frac{1}{6} \text{ as } n \to \infty \\&\text{Lower bound:} \\&\sum_{k=1}^n \frac{k}{3n^2 + n} = \frac{1}{n(3n + 1)} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2(3n + 1)} \\&\lim_{n \to \infty} \frac{n+1}{2(3n + 1)} = \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{2(3 + \frac{1}{n})} = \frac{1}{2 \cdot 3} = \frac{1}{6} \\&\text{By squeeze theorem, since both bounds approach } \frac{1}{6}: \\&\lim_{n \to \infty} \sum_{k=1}^n \frac{k}{3n^2 + k} = \frac{1}{6} \\&\text{Final answer:} \boxed{\frac{1}{6}}\end{aligned} $$

$$ \begin{aligned}&\text{Evaluate } \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{n + k}. \\&\text{The sum is:} \\&\sum_{k=1}^{n} \frac{1}{n + k} \\&\text{Rewrite the general term:} \\&\frac{1}{n + k} = \frac{1}{n \left(1 + \frac{k}{n}\right)} \\&\text{Thus, the sum becomes:} \\&\sum_{k=1}^{n} \frac{1}{n + k} = \sum_{k=1}^{n} \frac{1}{n} \cdot \frac{1}{1 + \frac{k}{n}} \\&\text{This resembles a Riemann sum for } f(x) = \frac{1}{1 + x} \text{ on } [0, 1], \\&\text{with } x_k = \frac{k}{n}, \ \Delta x = \frac{1}{n}: \\&\sum_{k=1}^{n} \frac{1}{n} \cdot \frac{1}{1 + \frac{k}{n}} \approx \int_0^1 \frac{1}{1 + x} \, dx \\&\text{Compute the integral:} \\&\int_0^1 \frac{1}{1 + x} \, dx = \left[ \ln(1 + x) \right]0^1 = \ln(2) - \ln(1) = \ln 2 \\&\text{Thus, the limit is likely:} \\&\lim{n \to \infty} \sum_{k=1}^{n} \frac{1}{n + k} = \ln 2\end{aligned} $$

$$ \begin{aligned}&\text{Derive the curvature formula from } \kappa = \left| \frac{d\mathbf{T}}{ds} \right| \text{ to } \kappa = \frac{|\mathbf{v} \times \mathbf{a}|}{|\mathbf{v}|^3}. \\&\text{Step 1: Understand the initial definition} \\&\text{Curvature is defined as:} \\&\kappa = \left| \frac{d\mathbf{T}}{ds} \right| \\&\text{where } \mathbf{T} = \frac{\mathbf{v}}{|\mathbf{v}|} \text{ is the unit tangent vector, } s \text{ is arc length,} \\&\mathbf{v} = \frac{d\mathbf{r}}{dt} \text{ is velocity, and } \mathbf{a} = \frac{d\mathbf{v}}{dt} \text{ is acceleration.} \\&\text{Step 2: Relate arc length } s \text{ to parameter } t \\&\frac{ds}{dt} = |\mathbf{v}|, \text{ so } \frac{d}{ds} = \frac{1}{|\mathbf{v}|} \cdot \frac{d}{dt} \\&\text{Thus:} \\&\frac{d\mathbf{T}}{ds} = \frac{1}{|\mathbf{v}|} \frac{d\mathbf{T}}{dt} \\&\kappa = \left| \frac{d\mathbf{T}}{ds} \right| = \frac{1}{|\mathbf{v}|} \left| \frac{d\mathbf{T}}{dt} \right| \\&\text{Step 3: Compute } \frac{d\mathbf{T}}{dt} \\&\mathbf{T} = \frac{\mathbf{v}}{|\mathbf{v}|}, \text{ let } v = |\mathbf{v}|. \text{ By quotient rule:} \\&\frac{d\mathbf{T}}{dt} = \frac{\frac{d\mathbf{v}}{dt} v - \mathbf{v} \frac{dv}{dt}}{v^2} \\&\text{Since } \mathbf{a} = \frac{d\mathbf{v}}{dt}, \text{ compute } \frac{dv}{dt}: \\&v = (\mathbf{v} \cdot \mathbf{v})^{1/2}, \ \frac{dv}{dt} = \frac{\mathbf{v} \cdot \mathbf{a}}{|\mathbf{v}|} \\&\text{So:} \\&\frac{d\mathbf{T}}{dt} = \frac{\mathbf{a} |\mathbf{v}| - \mathbf{v} \frac{\mathbf{v} \cdot \mathbf{a}}{|\mathbf{v}|}}{|\mathbf{v}|^2} = \frac{\mathbf{a} |\mathbf{v}|^2 - \mathbf{v} (\mathbf{v} \cdot \mathbf{a})}{|\mathbf{v}|^3} \\&\kappa = \frac{1}{|\mathbf{v}|} \cdot \frac{|\mathbf{a} |\mathbf{v}|^2 - \mathbf{v} (\mathbf{v} \cdot \mathbf{a})|}{|\mathbf{v}|^3} = \frac{|\mathbf{a} |\mathbf{v}|^2 - \mathbf{v} (\mathbf{v} \cdot \mathbf{a})|}{|\mathbf{v}|^4} \\&\text{Step 4: Simplify using the cross product} \\&\text{Consider } \mathbf{a} |\mathbf{v}|^2 - \mathbf{v} (\mathbf{v} \cdot \mathbf{a}). \text{ Compute its magnitude:} \\&\left| \mathbf{a} |\mathbf{v}|^2 - \mathbf{v} (\mathbf{v} \cdot \mathbf{a}) \right|^2 = \left( \mathbf{a} |\mathbf{v}|^2 - \mathbf{v} (\mathbf{v} \cdot \mathbf{a}) \right) \cdot \left( \mathbf{a} |\mathbf{v}|^2 - \mathbf{v} (\mathbf{v} \cdot \mathbf{a}) \right) \\&= |\mathbf{a}|^2 |\mathbf{v}|^4 - 2 (\mathbf{v} \cdot \mathbf{a})^2 |\mathbf{v}|^2 + (\mathbf{v} \cdot \mathbf{a})^2 |\mathbf{v}|^2 \\&= |\mathbf{a}|^2 |\mathbf{v}|^4 - (\mathbf{v} \cdot \mathbf{a})^2 |\mathbf{v}|^2 \\&\text{Use } |\mathbf{v} \times \mathbf{a}|^2 = |\mathbf{v}|^2 |\mathbf{a}|^2 - (\mathbf{v} \cdot \mathbf{a})^2: \\&= |\mathbf{v}|^2 \left( |\mathbf{a}|^2 |\mathbf{v}|^2 - (\mathbf{v} \cdot \mathbf{a})^2 \right) = |\mathbf{v}|^2 |\mathbf{v} \times \mathbf{a}|^2 \\&\text{Thus:} \\&\left| \mathbf{a} |\mathbf{v}|^2 - \mathbf{v} (\mathbf{v} \cdot \mathbf{a}) \right| = |\mathbf{v}| |\mathbf{v} \times \mathbf{a}| \\&\kappa = \frac{|\mathbf{v}| |\mathbf{v} \times \mathbf{a}|}{|\mathbf{v}|^4} = \frac{|\mathbf{v} \times \mathbf{a}|}{|\mathbf{v}|^3} \\&\text{Final answer:} \\&\boxed{\kappa = \frac{|\mathbf{v} \times \mathbf{a}|}{|\mathbf{v}|^3}}\end{aligned} $$